Partial answer
Note that, for $n\geq9$, it is easy to show that $S(n)$ always has $1817$ as last digits. This fact eliminates the posibility of $S(n)$ being some even perfect power. For odd integers, note that:
perfect powers of integers having $1$ as last digit also have $1$ as last digit, so $S(n)$ can not be a perfect power of some odd integer having $1$ as last digit.
perfect powers of integers having $3$ as last digit have $3,9,7,1$ as last digits, so $S(n)$ could be a perfect power of some odd integer having $3$ as last digit only for powers $P$ of the form $P^{4k+3}$. It would remain to show that there is not any power of some odd integer having $3$ as last digit of the form $P^{4k+3}$ that could end in $1817$.
perfect powers of integers having $5$ as last digit also have $5$ as last digit, so $S(n)$ can not be a perfect power of some odd integer having $5$ as last digit.
perfect powers of integers having $7$ as last digit have $7,9,3,1$ as last digits, so $S(n)$ could be a perfect power of some odd integer having $7$ as last digit only for powers $P$ of the form $P^{4k+1}$. It would remain to show that there is not any power of some odd integer having $7$ as last digit of the form $P^{4k+1}$ that could end in $1817$.
perfect powers of integers having $9$ as last digit have $9,1$ as last digits, so $S(n)$ can not be a perfect power of some odd integer having $9$ as last digit.
Indeed, the "fixed" last digits of $S(n)$ grow as $n$ grows, so if there exists a counterexample for the pending cases, maybe other ones with more fixed ending digits could work.
EDIT
Running a Python program, I have been checking all odd integers less than $10001$ having $3$ or $7$ as last digit, and powers of the form $4k+3$ or $4k+1$ respectively less than $10001$.
I have been able to check that there are both powers of odd integers having $3$ as last digit of the form $P^{4k+3}$, and powers of odd integers having $7$ as last digit of the form $P^{4k+3}$, that end in $1817$. For instance, $73^{443}$ or $137^{57}$, although the program showed at least fourteen counterexamples before I stopped it.
I have found the powers $137^{2057}$, $153^{1399}$, $297^{929}$, $313^{1587}$, and $473^{2403}$ before I stopped the program, all of them ending in $51817$, the last five digits of $S(n)$ for $n\geq 14$
I have found the power $777^{653}$ ending in $851817$, the last six digits of $S(n)$ for $n\geq 14$, before I stopped the program.
Therefore, although the suitable powers get scarcer, the "fixed" last digits approach does not seem so promising as it could, or at least some additional insight would be needed. It seems that the smallest powers ending in some given fixed digits of $S(n)$ are quite bigger than $S(n)$, so proving that this holds would be a possible line of attack.
EDIT 2
Noting that $\lim_{n\to\infty} \frac{n!^2}{S(n)}=1$, we have that $S(n)<n^{2n}$; hence, there is an important restriction on the size of the possible perfect powers. Proving a minimum size of the powers generating the "fixed" digits greater than $n^{2n}$ could be a way to solve the problem.
Note also that the number of "fixed" digits of $S(n)$ seems to be always near to $\frac{n}{2}$.
Best Answer
Per Sil's comment
it is sufficient to check $n \leq 11$. One can find small prime factors $p$ for $2 \leq n \leq 9$ and $n=11$ with the property that $p^2$ does not divide $S_n$. A direct PARI/GP computation for the case $n=10$ shows that $S_{10}$ is not square. Thus we can conclude that there is no perfect power for $n\ge 2$, and a possible prime (other than $S_2=5$) occurs only for $n=10$.