Let $X=(X_1,\ldots,X_n)$, where $X_i \sim N(0,1)$ are iid.
I'm looking for a result (and a proof outline) on the concentration of the max abs value of these Gaussians, $\|X\|_\infty$. That is, some result of the form $P(\bigl | \|X\|_\infty -\sqrt{2\log (2n)}\bigr |>t)<o(t)$, where $o(t)$ is any reasonable function that goes to $0$ as $t$ gets large.
I know these results: $E \|X\|_\infty \leq \sqrt{2 \log (2n)}$, $P(\|X\|_\infty \geq \sqrt{2 \log (2n)}+t)\leq 2\exp(-t^2 /2)$, which seems to be the "right tail" of the result I'm looking for.
So you just need to show that $P(\|X\|_\infty \leq \sqrt{2 \log (2n)}-\epsilon)$ is small. This is easy because $$ P\left(\|X\|_\infty \leq a_n\right) = P\left(|X_1|\leq a_n\right)^n = (2\Phi(a_n) - 1)^n\le \left(1-2\frac{a_n}{\sqrt{2\pi}(a_n^2 + 1)}e^{-a_n^2/2}\right)^n. $$ Plugging $a_n = \sqrt{2\log (2n)-\delta}$,
$$ P\left(\|X\|_\infty \leq a_n\right) \le \left(1-\frac{\sqrt{2\log (2n)-\delta}}{\sqrt{2\pi}n(2\log (2n)-\delta + 1)}e^{\delta/2}\right)^n\\ \le \left(1-\frac{1}{\sqrt{2\pi}n(\sqrt{2\log (2n)-\delta} + 1)}e^{\delta/2}\right)^n\\ \le \left(1-\frac{e^{\delta/2}}{\sqrt{2\pi}n(\sqrt{2\log (2n)} + 1)}\right)^n\\ \le \exp\left\{-\frac{e^{\delta/2}}{\sqrt{2\pi}(\sqrt{2\log (2n)}+ 1)}\right\}, $$ So you can take $\delta = K\log\log n$ with some $K$ large enough in order to make this small. You might be disappointed by the fact that this goes to infinity. This is not so bad as in fact the corresponding $\epsilon$ is of order $$ \frac{c\log \log n}{\sqrt{\log n}} $$ and does go to zero. If you want to have it fixed, you will get even some exponentially small estimates for probability.