Let $p$ be an odd prime and let $R=\Lambda_{\mathbb{F}_p}[x_1,\dots,x_n]$ be the exterior algebra on $n$ generators over the finite field with $p$ elements. This is a graded-commutative ring.

Is there a formula for the number of homogeneous degree $m$ elements in $R$, which square to zero?

More specifically, if we call this number $c_{p,n,m}$, I have some obscure reason to conjecture that it is of the form $$ c_{p,n,m} = \sum_{i=0}^{l} a_i p^{f_i(n)} $$

For some fixed rational numbers $a_i$ and integer valued (rational) polynomials $f_i$. In particular, this would imply that for a fixed $n$, this number is a polynomial in $p$.

Is it known to be true (or false)?

In the simplest case $m=2$, the problem reduces to counting anti-symmetric matrices of rank 2, which has a closed form formula confirming the conjecture. The next interesting case is $m=4$ (the odd ones being trivial), for which I already have no idea what the answer is.

###Edit: (some more details on the $m=2$ case) Since $p$ is odd, there is a 1-1 correspondence between 2-forms and anti-symmetric matrices. Using the canonical form for 2-forms

$$\omega = x_1\wedge x_2+x_3\wedge x_4 +...+ x_{2k-1}\wedge x_{2k}$$

we see that $2k$ is the rank of the associated matrix, but $k$ is also the unique number such that $\omega^k \ne 0$ and $\omega^{k+1}=0$. Thus, we are reduced to counting anti-symmetric matrices of rank 2. The general problem of counting anti-symmetric matrices over $\mathbb{F}_q$ by rank is solved for example in prop. 3.8 of this paper in terms of q-factorials. For rank 2 (to which we need to add 1 because of $\omega=0$) the answer is $$\frac{(q^n-1)(q^{n-1}-1)}{q^2-1}$$.

Unfortunately, non of this has a straightforward generalization for $m>2$ ...