As you mention in your update, you have a general answer, but if you want a concrete answer for the low-dimensional integral cohomology of $G = \operatorname{GL}(3,2)$ (or any other finite group!), you can use GAP, specifically the hap package. The following code computes $H^i(G, \mathbf{Z})$ for $i \leq 10$:
LoadPackage("hap");
G = GL(3,2);
List([1..10], x->GroupCohomology(G, x));
This will output:
[ [ ], [ ], [ 2 ], [ 4, 3 ], [ ], [ 2, 2, 7 ], [ 2 ], [ 4, 3 ], [ 2, 2 ], [ 2, 2 ] ]
which is a list of the abelian invariants of the cohomology groups $H^i(G, \mathbf{Z})$ for $1 \leq i \leq 10$, i.e. the first ten integral cohomology groups are $1, 1, C_2, C_4 \times C_3, 1, C_2 \times C_2 \times C_7, C_2, C_4 \times C_3, C_2 \times C_2$ and $C_2 \times C_2$ (for readability I here use $C_n = \mathbf{Z}/n\mathbf{Z}$).
If you want to compute the mod $p$ cohomology, and don't feel like using the above with the universal coefficient theorem by hand, you can compute this just as easily by using GroupCohomology(G, x, p). When $p=2$, the sequence of exponents of the first ten cohomology groups (together with $H^0(G, \mathbf{Z}/2\mathbf{Z})$) is
$$
1, 0, 1, 2, 1, 2, 3, 2, 3, 4, 3
$$
i.e.g., $H^3(G, \mathbf{Z}/2\mathbf{Z}) \cong (\mathbf{Z}/2\mathbf{Z})^2$. This sequence (of course) agrees with the generating function in @DaveBenson's answer, which expands as:
$$
\frac{1+t^3}{(1-t^2)(1-t^3)} = 1 + \quad + t^2 + 2 t^3 + t^4 + 2 t^5 + 3 t^6 + 2 t^7 + 3 t^8 + 4 t^9 + 3 t^{10} + ...
$$
