I assume that by maximal you mean with respect to inclusion. Then the answer is no. Consider the following counterexample on the real line:
Let $\mathcal{B}_\epsilon := \epsilon\mathbb{Z}$ and $\widetilde{\mathcal{B}}_\epsilon := 3/2 \epsilon \mathbb{Z}$. Both sets are maximal with respect to inclusion for any $\epsilon$, but for the corresponding measures you will get $m_\epsilon \rightharpoonup \frac{1}{2} \mathcal{H}^1$ and $\widetilde{m}_\epsilon \rightharpoonup \frac{1}{3} \mathcal{H}^1$. So if you have a third sequence $\widehat{\mathcal{B}}_\epsilon$ that alternates between the two, the corresponding measures will not converge.
What you should be able to get though by compactness is a weakly converging subsequence to something of the form $a(x)d\mathcal{H}^n$, with constants $0<c_1 < a(x) < c_2 <\infty$ corresponding to the optimal and "worst" sphere packing in the respective dimension.