*(I'm operating under the assumption that you want an upper bound that holds for all such lattices. If you have a specific lattice that interests you, then there are potentially other methods to try.)*

For an easy bound, you can just use basic packing arguments to see that no rank $n$ lattice can have more than, say, $(2M+1)^n$ points of length at most $M \cdot \lambda_1(L)$, where $\lambda_1(L)$ is the length of the shortest lattice vector, which for you is at least $1$. In particular, assume without loss of generality that the lattice lies in $\mathbb{R}^n$ (by viewing the lattice as embedded in its span), and place an $n$-dimensional ball of radius $\lambda_1(L)/2$ around each lattice vector. Notice that these balls are disjoint and that they all lie in a ball of radius $M+1/2$. Therefore, the total volume of the small balls must be at most the volume of the larger ball, which immediately yields the bound.

For stronger bounds with a much more difficult proof, first notice that the densest such lattice `should'' simply be $\mathbb{Z}^n$. This is morally true because $\mathbb{Z}^n$ has the minimal determinant amongst all such lattices. Since a`

typical'' set of volume $V$ in the subspace spanned by your lattice has $V/\det(L)$ points, the lattice with smallest determinant tends to have more points inside of it. E.g., this provably holds as $M \to \infty$.

I'm not sure if $\mathbb{Z}^n$ is *exactly* the densest such lattice for all $n,m, M$ (some related conjectures are false). But, we know that something close to this is true. In particular, in work with Regev, we showed that no lattice $L \subseteq \mathbb{R}^n$ whose sublattices $L' \subseteq L$ all satisfy $\det(L') \geq 1$ has ``significantly'' more points than $\mathbb{Z}^n$. Your lattices clearly satisfy this constraint, so the bound applies. Here's the precise statement, and there are of course more details in the paper:

I hope that the notation is clear. We write $rB_2^n$ for an $n$-dimensional ball of radius $n$, and you should think of our $\mathcal{L}$ as your lattice embedded in the subspace that it spans.

(You can almost certainly get tighter bounds for the specific class of rank $n$ lattices that are sublattices of $\mathbb{Z}^m$ (the restriction on the coordinates of the basis seems unlikely to matter much), but this might be sufficient for you.)