In the meantime I think I solved it. Here's my solution:
Let $\mu_k$ denote the $k$th central momentum of $X_i$, i.e, $\mu_k=\mathbb{E}((X_i-\mu)^k)$, and $Z_i\equiv X_i-\mu$ for all $i$. Thus $\mathbb{E}(Z_i)=0$. Since$$
\mathbb{V}(S_n^2)=\mathbb{E}(S_n^4)-(\mathbb{E}(S_n^2))^2=\mathbb{E}(S_n^4)-\sigma^4,
$$we derive an expression of $\mathbb{E}(S_n^4)$ in terms of $n$ and the moments. We can rewrite $S_n^2$ as\begin{align*}
S_n^2=\frac{n\sum_{i=1}^n Z_i^2-(\sum_{i=1}^nZ_i)^2}{n(n-1)}.
\end{align*}
Squaring leads to\begin{align*}
S_n^4=\frac{n^2(\sum_{i=1}^nZ_i^2)^2-2n(\sum_{i=1}^nZ_i^2)(\sum_{i=1}^nZ_i)^2+(\sum_{i=1}^nZ_i)^4}{n^2(n-1)^2}
\end{align*}
and so\begin{align*}
\mathbb{E}(S_n^4)=\frac{n^2\mathbb{E}((\sum_{i=1}^nZ_i^2)^2)-2n\mathbb{E}\left((\sum_{i=1}^nZ_i^2)(\sum_{i=1}^nZ_i)^2 \right)+\mathbb{E}((\sum_{i=1}^n Z_i)^4)}{n^2(n-1)^2}.
\end{align*}
Since $Z_1,...,Z_n$ are independent, we have that, for distinct $i,j,k$,\begin{align*}
\mathbb{E}(Z_iZ_j)=0,\hspace{5mm}\mathbb{E}(Z_i^3Z_j)=0,\hspace{5mm}\mathbb{E}(Z_i^2Z_jZ_k)=0
\end{align*}
and\begin{align*}
E(Z_i^2Z_j^2)=\mu_2^2=\sigma^4,\hspace{5mm}\mathbb{E}(Z_i^4)=\mu_4.
\end{align*}
Then, with simple algebraic simplifications, it can be shown that the following holds\begin{align*}
\mathbb{E}((\sum_{i=1}^nZ_i^2)^2)&=n\mu_4+n(n-1)\sigma^4,\\
\mathbb{E}\left((\sum_{i=1}^nZ_i^2)(\sum_{i=1}^nZ_i)^2 \right)&=n\mu_4+n(n-1)\sigma^4,\\
\mathbb{E}((\sum_{i=1}^n Z_i)^4)&=n\mu_4+3n(n-1)\sigma^4.
\end{align*}
Substituting these into the expansion of $\mathbb{E}(S_n^4)$ and simplifying leads to\begin{align*}
\mathbb{E}(S_n^4)=\frac{(n-1)\mu_4+(n^2-2n+3)\sigma^4}{n(n-1)}
\end{align*}
and so\begin{align}\label{var}
\mathbb{V}(S_n^2)=\mathbb{E}(S_n^4)-\sigma^4=\frac{(n-1)\mu_4+(n^2-2n+3)\sigma^4}{n(n-1)}-\sigma^4=\frac{\mu_4}{n}-\frac{\sigma^4(n-3)}{n(n-1)}.
\end{align}
