Consider the Sylow p-subgroup of $\mathrm{GL}_{n}(\mathbb{F% }_{p})$ by taking the group of upper triangular matrices with ones along the diagonal, namely $UT(n,\mathbb{F}_{p})$. $\mathrm{GL}_{n}(\mathbb{F}_{p})$ has $n_{p}=\prod_{k=2}^{n}\frac{p^{k}-1% }{p-1}$ Sylow $p$-subgroups, of which $UT(n,\mathbb{F}_{p})$ is only one. Suppose we have $e_{p^{m}}$ elementary abelian p-subgroups of rank $m$ in $UT(n,\mathbb{F}_{p})$. By Sylow's Theorem, every elementary abelian p-subgroup of $\mathrm{GL}_{n}(\mathbb{F}_{p})$ is conjugate to an elementary abelian p-subgroup of $UT(n,\mathbb{F}_{p})$. So, what reason that prevents the number of elementary abelian p-subgroups of rank $m$ in $\mathrm{GL}_{n}(\mathbb{F% }_{p})$ to be $n_{p}\times e_{p^{m}}$?.
Thank you in advance!
As Brauer Suzuki suggested, you are overcounting. To see this consider the following example with $n=4$. The set of matrices matching the pattern $$ \left(\begin{array}{cccc}1&0&*&*\\0&1&*&*\\0&0&1&0\\0&0&0&1\end{array}\right) $$ form an elementary abelian $p$-subgroup $E$ of rank four. You see that $E$ is normalized by e.g. the permutation matrices $\rho(\alpha)$ and $\rho(\beta)$ representing the 2-cycles $\alpha=(12)$ and $\beta=(34)$.
As neither $\rho(\alpha)$ nor $\rho(\beta)$ normalizes $UT(4)$ this means that in addition to $UT(4)$, the group $E$ appears as a subgroup of the Sylow $p$-subgroup $UT(4)^{\rho(\alpha)}$ of matrices of the shape $$ \left(\begin{array}{cccc}1&0&*&*\\*&1&*&*\\0&0&1&*\\0&0&0&1\end{array}\right) $$ as well as a subroup of $UT(4)^{\rho(\beta)}$ of matrices of the shape $$ \left(\begin{array}{cccc}1&*&*&*\\0&1&*&*\\0&0&1&0\\0&0&*&1\end{array}\right). $$