From considerations above, I would guess that, when $n=o(\sigma(n))$, $T^{\star}(n)=\mu+\tfrac{\sigma(n)}{\mu}$ satisfies $$\lim_n\mathbb{P}(X_{max}\ge T^{\star}(n))=1,$$ for any array $X^n_i$. If $T(n)>T^{\star}(n)$ the example given in the comments is a counterexample to $$\lim_n\mathbb{P}(X_{max}\ge T(n))=1.$$

From considerations above, I would guess that, when $n=o(\sigma(n))$, $T^{\star}(n)=\mu+\tfrac{\sigma(n)}{\mu}$ satisfies $$\lim_n\mathbb{P}(X_{max}\ge T^{\star}(n))=1,$$ for any array $X^n_i$ meeting the Bullmoose's conditions. If $T(n)>T^{\star}(n)$ the example given in the comments is a counterexample to $$\lim_n\mathbb{P}(X_{max}\ge T(n))=1.$$