I will sharpen my earlier comment a little. The number of such matrices is divisible by $(p-1)^{2n-2}.$ Let $T$ be the group of invertible diagonal matrices, and let $\mathcal{M}$ denote the set of invertible matrices of full rank ( though not necessarily determinant 1).
Then $M \to TMU^{-1}$ defines an action of $T \times T$ on $\mathcal{M},$ and the only pairs $(T,U)$ which fix any element are those with $T = U$ both scalar. Hence $(T \times T)/{\rm diag} (S)$ acts semi-regularly on $\mathcal{M},$ where $S$ is the subgroup of $T$ consisting of scalars. Thus $|\mathcal{M}|$ is divisible by $(p-1)^{2n-1}$, and hence, as user74230 notes in comments, the number of matrices in $\mathcal{M}$ which have determinant $1$ is a multiple of $(p-1)^{2n-2}.$ 

I also remark that the general problem looks quite messy to me for $n >2.$ Having chosen the first $k$ linearly independent rows with all non-zero entries, we have $(p-1)^{n} - f(k)$ ways to complete the $k+1$-st row, where $f(k)$ is the number of row vectors in the span of the first $k$-rows which have all non-zero entries. But it is not obvious (and is probably not true) that the number $f(k)$ is independent of the choice of the first $k$-rows. Also, the last row chosen may be multiplied by any non-zero scalar, and one and only one choice of scalar makes the whole matrix have determinant $1.$