Here is a quick and dirty argument. For $SL(3,3)$ we can check directly that there is no representation of dimension less than $11$ over a field of characteristic $\ne 3$. So we can assume that $n\geqslant 11$. On the other hand, the Sylow $3$-subgroup of $SL(n,3)$ has derived length at least $\log_2(n)$, whereas the Sylow $3$-subgroup of $SL(n,p)$ for $p>3$ has derived length at most $\log_3(n)+1$. If $\log_2(n)\leqslant \log_3(n)+1$ then $n\leqslant 2^{1/(1-\log_3(2))} < 7$. This contradiction shows that there is no $n\ge 3$ for which $SL(n,3)$ embeds in $SL(n,p)$ with $p>3$. I hope I haven't made any stupid mistakes here.

$\DeclareMathOperator\SL{SL}$Here is a quick and dirty argument. For $\SL(3,3)$ we can check directly that there is no representation of dimension less than $11$ over a field of characteristic $\ne 3$. So we can assume that $n\geqslant 11$. On the other hand, the Sylow $3$-subgroup of $\SL(n,3)$ has derived length at least $\log_2(n)$, whereas the Sylow $3$-subgroup of $\SL(n,p)$ for $p>3$ has derived length at most $\log_3(n)+1$. If $\log_2(n)\leqslant \log_3(n)+1$ then $n\leqslant 2^{1/(1-\log_3(2))} < 7$. This contradiction shows that there is no $n\ge 3$ for which $\SL(n,3)$ embeds in $\SL(n,p)$ with $p>3$. I hope I haven't made any stupid mistakes here.