Well, by Hamilton--Cayley, each matrix $A\in {\rm GL}(n,p)$ generates an at most $n$-dimensional subalgebra ${\mathbb F}_p[A]\subseteq M(n,p)$ thus containing at most $p^n-1$ nonzero elements. Hence the order of $A$ cannot exceed $p^n-1$.
On the other hand, consider a degree $n$ monic polynomial $P_n$ whose root is a generator $\xi$ of ${\mathbb F}_{p^n}^*$. Then a matrix with $P_n$ as its characteristic polynomial has order at least $p^n-1$ since $\xi$ is its eigenvalue.
ADDENDUM. if you wish the order to be the power of $p$, then the answer is $d=p^{\lceil \log_p n\rceil}$. Since the order of $A$ is divisible by the multiplicative orders of its eigenvalues, all the eigenvalues should be $1$. Hence the characteristic polynomial is $(x-1)^n$, so $A^d-I=(A-I)^d=0$.
On the other hand, if $A=I+J$ is the Jordan cell of size $n$ (with eigenvalue 1), then $A^{d/p}=I^{d/p}+J^{d/p}\neq I$, but $A^d=I+J^d=I$.
NB. The subgroup of all (upper-)unitriangular matrices is a Sylow $p$-subgroup in ${\rm GL}(n,p)$. So you may concentrate on it when looking at the elements of this kind.
