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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. Last edit: For completeness, I will give an argument for this case: A maximal parabolic $P$ of ${\rm SL}(3,3)$ is a semidirect product of an elementary Abelian subgroup of order $9$ with ${\rm GL}(2,3).$ In particular, $P$ contains a Frobenius group of order $72$ with kernel of order $9$. It follows that $P$ has no faithful complex character of degree less than $8$ (in fact, it does have two faithful irreducible characters of degree $8$). Since $P$ is solvable, it follows by the Fong-Swan theorem that $P$ has no faithful representation of degree less than $8$ over any field of characteristic $p > 3.$ Hence $P$ does not embed as a subgroup of ${\rm SL}(3,p)$ when $p >3.$

$\DeclareMathOperator\SL{SL}$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 $\SL(n,p)$ 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 $\SL(n,3)$ 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 $\SL(n,p),$ 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 $\SL(3,p)$ may contain extra-special subgroups of order $27$ in that case. Last edit: For completeness, I will give an argument for this case: A maximal parabolic $P$ of $\SL(3,3)$ is a semidirect product of an elementary Abelian subgroup of order $9$ with $\operatorname{GL}(2,3)$. In particular, $P$ contains a Frobenius group of order $72$ with kernel of order $9$. It follows that $P$ has no faithful complex character of degree less than $8$ (in fact, it does have two faithful irreducible characters of degree $8$). Since $P$ is solvable, it follows by the Fong–Swan theorem that $P$ has no faithful representation of degree less than $8$ over any field of characteristic $p > 3$. Hence $P$ does not embed as a subgroup of $\SL(3,p)$ when $p >3$.

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.

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. Last edit: For completeness, I will give an argument for this case: A maximal parabolic $P$ of ${\rm SL}(3,3)$ is a semidirect product of an elementary Abelian subgroup of order $9$ with ${\rm GL}(2,3).$ In particular, $P$ contains a Frobenius group of order $72$ with kernel of order $9$. It follows that $P$ has no faithful complex character of degree less than $8$ (in fact, it does have two faithful irreducible characters of degree $8$). Since $P$ is solvable, it follows by the Fong-Swan theorem that $P$ has no faithful representation of degree less than $8$ over any field of characteristic $p > 3.$ Hence $P$ does not embed as a subgroup of ${\rm SL}(3,p)$ when $p >3.$

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.

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.