Let me make Robert's answer a bit more concrete. The theory of skewsymmteric forms is much nicer than the symmetric one since they are classified by their rank for any field in any characteristic (including $p=2$). Now the stabilizer of the standard form $$ e_1\wedge e_2+\ldots+e_{2r-1}\wedge e_{2r} $$ is $Sp(2r)\times GL(n-2r)\ltimes U$. Thus the number of rank-2-skewsymmetric forms in $n$-space is $$ a(n,r)=\frac{|GL(n)|}{|Sp(2r)|\cdot|GL(n-2r)|\cdot|U|} $$ where $$ |GL(n)|=q^{\frac12 n(n-1)}\prod_{i=1}^n(q^i-1),\quad |Sp(2r)|=q^{r^2}\prod_{i=1}^r(q^{2i}-1),\quad |U|=q^{2r(n-2r)}. $$ Thus $$ a(n,r)=q^{r(r-1)}\prod_{i=1}^r(q^{2i-1}-1)\left[\matrix{n\\2r}\right]_q $$

Let me make Robert's answer a bit more concrete. The theory of skewsymmteric forms is much nicer than the symmetric one since they are classified by their rank for any field in any characteristic (including $p=2$). Now the stabilizer of the standard form $$ e_1\wedge e_2+\ldots+e_{2r-1}\wedge e_{2r} $$ is $Sp(2r)\times GL(n-2r)\ltimes U$. Thus the number of rank-$2r$-skewsymmetric forms in $n$-space is $$ a(n,r)=\frac{|GL(n)|}{|Sp(2r)|\cdot|GL(n-2r)|\cdot|U|} $$ where $$ |GL(n)|=q^{\frac12 n(n-1)}\prod_{i=1}^n(q^i-1),\quad |Sp(2r)|=q^{r^2}\prod_{i=1}^r(q^{2i}-1),\quad |U|=q^{2r(n-2r)}. $$ Thus $$ a(n,r)=q^{r(r-1)}\ \prod_{i=1}^r(q^{2i-1}-1)\ \left[\matrix{n\\2r}\right]_q $$