Is uniform convergence faster for low-entropy distributions?
First, let's use McDiarmid's inequality to conclude that
$$\mathbb{P}\left[|| \bar X - \mu ||_\infty \ge 
\mathbb{E}|| \bar X - \mu ||_\infty
+
\varepsilon \right] \le  e^{-2n\varepsilon^2},$$
so it remains to bound $\mathbb{E}|| \bar X - \mu ||_\infty$.
Using Jensen's inequality,
$$ (\mathbb{E}|| \bar X - \mu ||_\infty)^2\le \mathbb{E}|| \bar X - \mu ||_\infty^2
\le \sum_{j=1}^d\mathbb{E}(\bar X(j)-\mu(j))^2=
\sum_{j=1}^d\mathrm{Var}[\bar X(j)]=\frac1{n^2}\sum_{j=1}^d\mathrm{Var}[X(j)],
$$
whence
$$ \mathbb{E}|| \bar X - \mu ||_\infty\le
\frac1{n}\sqrt{\sum_{j=1}^d\mathrm{Var}[X(j)]}.
$$
For sufficiently small variances, this is an improvement over your bound.

Update. A different approach would be to introduce the Rademacher random variables $\sigma_i$ and to use the standard symmetrization argument to obtain
$$ \mathbb{E}|| \bar X - \mu ||_\infty
\le2\mathbb{E}\max_{j\in[d]}\frac1n\sum_{i=1}^n\sigma_i X_i(j), 
$$
where $X_i(j)$ indicates the $j$th bit of $X_i$. Now there must be a way to bound the latter Rademacher complexity using the entropy of $\mathcal{D}$, but I don't have anything off the top of my head. Massart's finite class lemma will give you a distribution-free bound of $O(\sqrt{\log(d)/n})$.