The idea I had in mind originally when I made the comment was more simple-minded. I'll deal with the case that $n$ is even for ease of exposition.
Each $3$-subgroup $S$ of ${\rm SL}(n,p)$ has an Abelian normal subgroup of index dividing the $3$-part of $n!$, so certainly dividing $3^{\frac{n-1}{2}}.$
A Sylow $3$-subgroup $T$ of ${\rm SL}(n,3)$ has order $3^{\frac{n(n-1)}{2}}$, and the largest Abelian subgroup of $T$ has order $3^{\frac{n^{2}}{4}},$ so $T$ has no Abelian subgroup of index less than $3^{\frac{n^{2}}{4} -\frac{n}{2}}.$
If $T$ embeds in ${\rm SL}(n,p),$ then we must have $\frac{n^{2}}{4} \leq \frac{n}{2} + \frac{n-1}{2} < n,$ so $n < 4$ (since $n$ is even, this forces $n=2).$
The argument for $n$ odd is similar.
Later edit: In fact, I think the case $n = 3$ and $p \equiv 1$ (mod $3$) needs a separate argument, since ${\rm SL}(3,p)$ may contain extra-special subgroups of order $27$ in that case.