Note that the set of available configurations is in fact a $\mathbb{F}_2$-vector space. In one dimension (that is, for $k=1$), it's easy to see that that vector space has dimension $n-1$: you can get all but one of the lattice points to be whatever you want, and then the last one is forced to be the sum of the ones that remain.

Call that space $V\subset\mathbb{F}_2^n$.

Now, in general, you're working with the $k$-th tensor power of $V$, and hence the dimension is $(n-1)^k$, and hence the number of configurations is $2^{(n-1)^k}$. Why's that? Well, the vector space of configurations, is generated by operations which are exactly tensor products of the generators of $V$.