[proofplan]
We expand each cumulant of $S_n$ by multilinearity and then use freeness to remove all mixed cumulants. Only the terms in which all entries come from the same variable remain, and identical distribution makes all surviving diagonal terms equal to the cumulant of $a_1$. This gives the exact scaling formula $\kappa_m(S_n,\dots,S_n)=n^{1-m/2}\kappa_m(a_1,\dots,a_1)$, from which the cases $m=1$, $m=2$, and $m\geq 3$ follow by evaluating the exponent.
[/proofplan]
[step:Expand the normalized cumulant by multilinearity]
Fix an integer $m \geq 1$. For each $n \in \mathbb{N}$, define the index set
\begin{align*}
I_m(n) = \{1,\dots,n\}^m.
\end{align*}
By multilinearity of the free cumulant functional $\kappa_m: \mathcal{A}^m \to \mathbb{C}$ in its $m$ arguments,
\begin{align*}
\kappa_m(S_n,\dots,S_n) = n^{-m/2}\sum_{(i_1,\dots,i_m)\in I_m(n)} \kappa_m(a_{i_1},\dots,a_{i_m}).
\end{align*}
Here the scalar factor $n^{-m/2}$ comes from the factor $n^{-1/2}$ in each of the $m$ arguments of $\kappa_m$.
[/step]
[step:Use freeness to keep only same-index cumulants]
Define the diagonal index set
\begin{align*}
D_m(n) = \{(k,\dots,k)\in I_m(n): 1 \leq k \leq n\}.
\end{align*}
Since the variables $(a_k)_{k\geq 1}$ are freely independent, every mixed free cumulant involving entries from at least two different freely independent subalgebras vanishes. Therefore, if $(i_1,\dots,i_m)\in I_m(n)\setminus D_m(n)$, then
\begin{align*}
\kappa_m(a_{i_1},\dots,a_{i_m}) = 0.
\end{align*}
Hence
\begin{align*}
\kappa_m(S_n,\dots,S_n) = n^{-m/2}\sum_{k=1}^{n}\kappa_m(a_k,\dots,a_k).
\end{align*}
[guided]
We now isolate the effect of freeness. The expansion from the previous step contains one term for every ordered $m$-tuple $(i_1,\dots,i_m)\in I_m(n)$. A term is called mixed here when not all of the indices $i_1,\dots,i_m$ are equal. Equivalently, the mixed terms are precisely those indexed by $I_m(n)\setminus D_m(n)$, where
\begin{align*}
D_m(n) = \{(k,\dots,k)\in I_m(n): 1 \leq k \leq n\}.
\end{align*}
The defining cumulant characterization of freeness says that mixed free cumulants of freely independent variables vanish. Since the variables $a_1,a_2,\dots$ are freely independent, every tuple $(i_1,\dots,i_m)$ using at least two distinct indices gives
\begin{align*}
\kappa_m(a_{i_1},\dots,a_{i_m}) = 0.
\end{align*}
Thus all nonzero contributions in the multilinear expansion must come from tuples of the form $(k,\dots,k)$ for some $1 \leq k \leq n$. Substituting this into the expansion gives
\begin{align*}
\kappa_m(S_n,\dots,S_n) = n^{-m/2}\sum_{k=1}^{n}\kappa_m(a_k,\dots,a_k).
\end{align*}
This is the key reduction: freeness changes a sum over $n^m$ possible index choices into a sum over only $n$ diagonal choices.
[/guided]
[/step]
[step:Use identical distribution to obtain the cumulant scaling formula]
Since the variables $(a_k)_{k\geq 1}$ are identically distributed and have moments of all orders, their one-variable free cumulants agree in every order. Define the scalar
\begin{align*}
c_m = \kappa_m(a_1,\dots,a_1).
\end{align*}
Then, for every $1 \leq k \leq n$,
\begin{align*}
\kappa_m(a_k,\dots,a_k) = c_m.
\end{align*}
Therefore
\begin{align*}
\kappa_m(S_n,\dots,S_n) = n^{-m/2}\sum_{k=1}^{n} c_m = n^{1-m/2}c_m.
\end{align*}
Equivalently,
\begin{align*}
\kappa_m(S_n,\dots,S_n) = n^{1-m/2}\kappa_m(a_1,\dots,a_1).
\end{align*}
[/step]
[step:Evaluate the scaling formula in orders one and two]
For $m=1$, the first free cumulant equals the expectation, so
\begin{align*}
\kappa_1(a_1) = \varphi(a_1).
\end{align*}
Since the variables are centered, $\varphi(a_1)=0$. The scaling formula gives
\begin{align*}
\kappa_1(S_n) = n^{1/2}\kappa_1(a_1) = 0.
\end{align*}
For $m=2$, the exponent is $1 - 2/2 = 0$, so the scaling formula gives
\begin{align*}
\kappa_2(S_n,S_n) = n^0\kappa_2(a_1,a_1) = \kappa_2(a_1,a_1).
\end{align*}
[/step]
[step:Let the normalization force all higher cumulants to zero]
Let $m \geq 3$ be fixed. Then
\begin{align*}
1 - \frac{m}{2} < 0.
\end{align*}
Since $a_1$ has moments of all orders, the scalar $\kappa_m(a_1,\dots,a_1)\in \mathbb{C}$ is finite and independent of $n$. Hence
\begin{align*}
\lim_{n\to\infty} n^{1-m/2}\kappa_m(a_1,\dots,a_1) = 0.
\end{align*}
Using the scaling formula, this is exactly
\begin{align*}
\lim_{n\to\infty}\kappa_m(S_n,\dots,S_n) = 0.
\end{align*}
This proves the asserted vanishing of all free cumulants of order $m\geq 3$ under the normalization $n^{-1/2}$.
[/step]
