My approach fails for too many things are heuristic, nothing new, but your assumption entails $n=o(\sigma(n))$, so at least this information is good news. Thanks for the nice question.

Since the asymptotic behavior of the maximum usually depends much more on the tail behavior than on the variance, a bounded variable should achieve the lowest possible maximum, if there exists such a thing: a $X_{max}$ that is stochastically minimal.

I assume here that the maximal variance for a random variable $X$ with given expectation $\mu$ and $a\le X\le b$ is achieved when $X\in\{a,b\}$ a.s..

Sorry, the parameter should be $(1+\tfrac{\sigma_n^2}{\mu^2})^{-1}$, and $\sigma_n^2=\sigma(n)$. Then $\mathbb{P}(X_{max}<\tfrac{\sigma_n^2+\mu^2}\mu)=(1+\tfrac{\mu^2}{\sigma_n^2})^{-n}$, if I did not do another mistake ...

Heuristically, it seems that we need to know more about the variance $\sigma_n^2$. For instance, if the $X_i^n$ have the smaller essential supremum possible, which I would guess is when $X_i^n$ is $\tfrac{\mu^2+\sigma_n^2}\mu$ times a Bernoulli random variable with parameter $(1+\tfrac{\mu^2}{\sigma_n^2})^{-1}$, then if $n=o(\sigma_n^2)$ the max is $\tfrac{\mu^2+\sigma_n^2}\mu$ with a probability close to one, which is my guess for a lower bound, in this case. But this is heuristic at many stages.