Do you really mean SL_n or GL_n? For GL_n this was proven by Hotta and Springer in "A Specialisation Theorem for ...". However I thought this map is already not surjective for all unipotent elements of SL_n.

It depends what you mean by natural. In Theorem 4.3.34 of James and Kerber's book "The Representation Theory of the Symmetric Group" a complete list of irreducible representations is obtained via Clifford theory. However this is probably not what you're looking for.

Note that as Geoff points out in the case of $\mathrm{GL}_n(q)$ this is quite easy to deduce from the order formula for $\mathrm{GL}_n(q)$, which is quite easy to work out without the theory of algebraic groups.

Thanks, once again, for answering my question. This is so simple, I feel like I really should have come across it before.

Sorry, I was in the process of changing an answer to the OP's original question before he changed it. The above now explains why $SL_2(p)$ has irreducible characters of degree $\frac{p-1}{2}$ and $\frac{p+1}{2}$.