For a subcollection $\mathcal S$ of $k$-element subsets of $[n]$, consider the random variable $Z_{\mathcal S} := \sup_{A \in \mathcal S}|X_A|$, where $X_A:=\sum_{i \in A}X_i$, and the $X_i$'s are iid from $N(0,1)$. Note that $X_A \sim N(0,k)$. Moreover, it is a standard computation that $$ \begin{split} \mathbb E Z_{\mathcal S} &= \mathbb E \sup_{A \in \mathcal S}|X_A| \le \sqrt{2k\log|\mathcal S|} < \infty,\\ \sigma_{\mathcal S}^2 &:= \sup_{A \in \mathcal S}\mathbb E|X_A|^2 = k < \infty. \end{split} $$ Therefore, noting that $Z_{\mathcal S} \ge \sup_{A \in \mathcal S}X_A$, the Borell-TIS ienquality gives
$$ \begin{split} \forall u \ge 0,\; \mathbb P\left(\sup_{A \in \mathcal S}X_A \ge \sqrt{2k\log|\mathcal S|} + u\right) &\le \mathbb P(Z_{\mathcal S} \ge \sqrt{2k\log|\mathcal S|} + u)\\ &\le \mathbb P(Z_{\mathcal S} \ge \mathbb EZ_{\mathcal S} + u)\\ &\le \exp(-u^2/(2\sigma_{\mathcal S}^2)) = \exp(-u^2/(2k)). \end{split} $$
To make things more interpretable, we do the the change of variable $t:=u/\sqrt{2k}$ to get $$ \mathbb P\left(\sup_{A \in \mathcal S}X_A \le \sqrt{2k}(\sqrt{\log|\mathcal S|} + t)\right) \le e^{-t^2},\forall t \ge 0. $$