I would take the standard cyclic resolution of $G = Z/nZ$:
$$
\dots \stackrel{1-t}\to Z[G] \stackrel{\sum t^i}\to Z[G] \stackrel{1-t}\to Z[G] \to Z \to 0,
$$
where $t$ is the generator of $G$, and then take the tensor square of two such --- this would give a resolution
$$
\dots \to Z[G_1\times G_2]^3 \stackrel{d_2}\to Z[G_1\times G_2]^2 \stackrel{d_1}\to Z[G_1\times G_2] \to Z \to 0,
$$
where $G_1 = G_2 = Z/nZ$ and the maps are given by
$$
d_1 = (1-t_1,1-t_2),
\qquad
d_2 = \left(\begin{array}{ccc}
\sum t_1^i & 1-t_2 & 0 \cr
0 & 1-t_1 & \sum t_2^i
\end{array}\right)
$$
($t_1$ and $t_2$ are the generators of $G_1$ and $G_2$ respectively).
I think you can use this for the calculations.