[proofplan]
We use the pointwise minor-arc estimate for the von Mangoldt exponential sum with two spare powers of $\log N$. The cubic integral is bounded by the minor-arc supremum of $|S_N|$ times the global second moment of $S_N$. The second moment is computed by finite Fourier orthogonality on $\mathbb T$, and the resulting sum of $\Lambda(n)^2$ is bounded by the elementary inequality $\Lambda(n)\le \log N$ for $n\le N$.
[/proofplan]
[step:Choose the logarithmic saving in the pointwise minor-arc estimate]
Fix $A>0$ and set
\begin{align*}
A_1:=A+2.
\end{align*}
The [Vinogradov Minor-Arc Estimate for the Von Mangoldt Exponential Sum][citetheorem:9083] applies to the von Mangoldt function $\Lambda:\mathbb N\to\mathbb R$, the exponential sum $S_N:\mathbb T\to\mathbb C$ from the statement, and the major arcs $\mathfrak M_N(Q,R)$ defined by the condition $\|\alpha-a/q\|_{\mathbb R/\mathbb Z}\le R/(qN)$ with $1\le q\le Q$ and $\gcd(a,q)=1$. Its quantifiers say that, for every prescribed saving exponent, one may choose a logarithmic exponent for $Q$ and $R$ so that the complementary minor arcs have the corresponding pointwise bound. Applying it with prescribed saving $A_1$ and then setting the theorem's width parameter $L$ equal to that cited parameter $R$, there exist constants $B=B(A)>0$, $C_1=C_1(A)>0$, and $N_1=N_1(A)\in\mathbb N$ with $N_1\ge 3$ such that, for every $N\ge N_1$, with $Q=(\log N)^B$, $L=(\log N)^B$, and $\mathfrak m=\mathbb T\setminus\mathfrak M_N(Q,L)$, one has
\begin{align*}
\sup_{\alpha\in\mathfrak m}|S_N(\alpha)|\le C_1\frac{N}{(\log N)^{A+2}}.
\end{align*}
[guided]
The proof needs one input from the circle method: a pointwise estimate for $S_N(\alpha)$ away from the major arcs. We deliberately ask for the stronger saving $A+2$, not merely $A$, because the second-moment estimate below will cost at most two powers of $\log N$.
Define the auxiliary exponent
\begin{align*}
A_1:=A+2.
\end{align*}
The [Vinogradov Minor-Arc Estimate for the Von Mangoldt Exponential Sum][citetheorem:9083] applies to the same map
\begin{align*}
S_N:\mathbb T\to\mathbb C,\quad \alpha\mapsto \sum_{1\le n\le N}\Lambda(n)e(n\alpha),
\end{align*}
and to major arcs $\mathfrak M_N(Q,R)$ defined by the same rational approximation condition as in the statement, namely $1\le q\le Q$, $\gcd(a,q)=1$, and $\|\alpha-a/q\|_{\mathbb R/\mathbb Z}\le R/(qN)$. The theorem's quantifiers allow an arbitrary prescribed logarithmic saving after choosing the exponent in $Q=R=(\log N)^B$ large enough. Therefore, with saving $A_1=A+2$ and with the present statement's width parameter $L$ identified with that parameter $R$, there are constants $B=B(A)>0$, $C_1=C_1(A)>0$, and $N_1=N_1(A)\in\mathbb N$ with $N_1\ge 3$ such that, whenever $N\ge N_1$ and $Q=(\log N)^B$, $L=(\log N)^B$, and $\mathfrak m=\mathbb T\setminus\mathfrak M_N(Q,L)$, we have
\begin{align*}
\sup_{\alpha\in\mathfrak m}|S_N(\alpha)|\le C_1\frac{N}{(\log N)^{A+2}}.
\end{align*}
This is exactly where the choice of $B$ is made: $B$ is chosen large enough for the minor-arc theorem with the stronger exponent $A+2$.
For this same $N$, the function $\alpha\mapsto |S_N(\alpha)|^3$ is continuous on $\mathbb T$, because $S_N$ is a finite trigonometric polynomial. Hence it is measurable and integrable with respect to $\mu_{\mathbb T}$. If $\mathfrak m=\varnothing$, then the integral over $\mathfrak m$ is $0$, so the desired estimate follows after increasing constants if necessary. Otherwise $\mathfrak m$ is nonempty, and for every $\alpha\in\mathfrak m$ we have
\begin{align*}
|S_N(\alpha)|^3\le \left(\sup_{\beta\in\mathfrak m}|S_N(\beta)|\right)|S_N(\alpha)|^2.
\end{align*}
Integrating this pointwise inequality over $\mathfrak m$ and then enlarging the nonnegative integral from $\mathfrak m$ to $\mathbb T$ gives
\begin{align*}
\int_{\mathfrak m}|S_N(\alpha)|^3\,d\mu_{\mathbb T}(\alpha)\le \left(\sup_{\beta\in\mathfrak m}|S_N(\beta)|\right)\int_{\mathbb T}|S_N(\alpha)|^2\,d\mu_{\mathbb T}(\alpha).
\end{align*}
We now compute the second moment. For each integer $r\in\mathbb Z$, normalized Haar measure on $\mathbb T$ satisfies
\begin{align*}
\int_{\mathbb T}e(r\alpha)\,d\mu_{\mathbb T}(\alpha)=1
\end{align*}
when $r=0$, and
\begin{align*}
\int_{\mathbb T}e(r\alpha)\,d\mu_{\mathbb T}(\alpha)=0
\end{align*}
when $r\ne 0$. Since $S_N$ is a finite sum, expansion and finite linearity of the integral give
\begin{align*}
\int_{\mathbb T}|S_N(\alpha)|^2\,d\mu_{\mathbb T}(\alpha)=\sum_{1\le m\le N}\sum_{1\le n\le N}\Lambda(m)\Lambda(n)\int_{\mathbb T}e((m-n)\alpha)\,d\mu_{\mathbb T}(\alpha).
\end{align*}
The orthogonality relation leaves exactly the diagonal terms $m=n$, so
\begin{align*}
\int_{\mathbb T}|S_N(\alpha)|^2\,d\mu_{\mathbb T}(\alpha)=\sum_{1\le n\le N}\Lambda(n)^2.
\end{align*}
For $1\le n\le N$, the von Mangoldt function satisfies $0\le \Lambda(n)\le \log N$. Therefore
\begin{align*}
\sum_{1\le n\le N}\Lambda(n)^2\le \sum_{1\le n\le N}(\log N)^2=N(\log N)^2.
\end{align*}
Combining the pointwise minor-arc estimate with this second-moment bound gives
\begin{align*}
\int_{\mathfrak m}|S_N(\alpha)|^3\,d\mu_{\mathbb T}(\alpha)\le C_1\frac{N}{(\log N)^{A+2}}N(\log N)^2.
\end{align*}
Thus
\begin{align*}
\int_{\mathfrak m}|S_N(\alpha)|^3\,d\mu_{\mathbb T}(\alpha)\le C_1\frac{N^2}{(\log N)^A}.
\end{align*}
Taking $C_A:=C_1(A)$ and $N_A:=\max\{N_1(A),2\}$ proves the claimed estimate.
[/guided]
[/step]
[step:Reduce the cubic minor-arc integral to a global second moment]
For $N\ge N_1$, the function $\alpha\mapsto |S_N(\alpha)|^3$ is continuous on $\mathbb T$, hence measurable and integrable with respect to $\mu_{\mathbb T}$. The set $\mathfrak M_N(Q,L)$ is a finite union of closed balls in $\mathbb T$, so $\mathfrak m=\mathbb T\setminus\mathfrak M_N(Q,L)$ is Borel measurable. If $\mathfrak m=\varnothing$, then the integral over $\mathfrak m$ is $0$, and the desired estimate follows. We may therefore assume $\mathfrak m\ne\varnothing$. Since $\mathfrak m\subseteq\mathbb T$, we have
\begin{align*}
\int_{\mathfrak m}|S_N(\alpha)|^3\,d\mu_{\mathbb T}(\alpha)\le \left(\sup_{\alpha\in\mathfrak m}|S_N(\alpha)|\right)\int_{\mathfrak m}|S_N(\alpha)|^2\,d\mu_{\mathbb T}(\alpha).
\end{align*}
Enlarging the domain of integration from $\mathfrak m$ to $\mathbb T$ gives
\begin{align*}
\int_{\mathfrak m}|S_N(\alpha)|^3\,d\mu_{\mathbb T}(\alpha)\le \left(\sup_{\alpha\in\mathfrak m}|S_N(\alpha)|\right)\int_{\mathbb T}|S_N(\alpha)|^2\,d\mu_{\mathbb T}(\alpha).
\end{align*}
[/step]
[step:Compute the second moment by finite Fourier orthogonality]
For integers $r\in\mathbb Z$, normalized Haar measure on $\mathbb T$ satisfies
\begin{align*}
\int_{\mathbb T}e(r\alpha)\,d\mu_{\mathbb T}(\alpha)=1
\end{align*}
when $r=0$, and
\begin{align*}
\int_{\mathbb T}e(r\alpha)\,d\mu_{\mathbb T}(\alpha)=0
\end{align*}
when $r\ne 0$. Since $S_N$ is a finite trigonometric polynomial, expanding the square and using finite additivity of the integral gives
\begin{align*}
\int_{\mathbb T}|S_N(\alpha)|^2\,d\mu_{\mathbb T}(\alpha)=\sum_{1\le m\le N}\sum_{1\le n\le N}\Lambda(m)\Lambda(n)\int_{\mathbb T}e((m-n)\alpha)\,d\mu_{\mathbb T}(\alpha).
\end{align*}
The orthogonality relation removes all terms with $m\ne n$, so
\begin{align*}
\int_{\mathbb T}|S_N(\alpha)|^2\,d\mu_{\mathbb T}(\alpha)=\sum_{1\le n\le N}\Lambda(n)^2.
\end{align*}
[/step]
[step:Bound the second moment by an elementary logarithmic estimate]
For every integer $n$ with $1\le n\le N$, the von Mangoldt function satisfies
\begin{align*}
0\le \Lambda(n)\le \log N.
\end{align*}
Therefore
\begin{align*}
\sum_{1\le n\le N}\Lambda(n)^2\le \sum_{1\le n\le N}(\log N)^2=N(\log N)^2.
\end{align*}
Combining this with the second-moment identity gives
\begin{align*}
\int_{\mathbb T}|S_N(\alpha)|^2\,d\mu_{\mathbb T}(\alpha)\le N(\log N)^2.
\end{align*}
[/step]
[step:Combine the bounds and absorb the spare logarithms]
For every $N\ge N_1$, the estimates above give
\begin{align*}
\int_{\mathfrak m}|S_N(\alpha)|^3\,d\mu_{\mathbb T}(\alpha)\le C_1\frac{N}{(\log N)^{A+2}}N(\log N)^2.
\end{align*}
Thus
\begin{align*}
\int_{\mathfrak m}|S_N(\alpha)|^3\,d\mu_{\mathbb T}(\alpha)\le C_1\frac{N^2}{(\log N)^A}.
\end{align*}
Taking $C_A:=C_1(A)$ and $N_A:=N_1(A)$ proves the claimed estimate, with $N_A\ge 3$ already arranged in the first step.
[/step]
