Given a random vector
\begin{equation} x=(x_1, \ldots, x_n) \end{equation}
with independent and identically distributed entries $x_i \sim \mathcal{N}(0,\sigma^2)$, I would like to find a lower bound $f(n)$
\begin{equation} \mathbb{E}[\|x\|^2_{\infty}] \geq f(n) \end{equation}
which is reasonably tight. I know that the following equality for the non squared norm holds when $\sigma^2 =1$:
\begin{equation} E(\|x\|_\infty)=\int_0^\infty(1-(2\Phi(x)-1)^n)dx, \end{equation}
where $\Phi$ is the CDF of $\mathcal{N}(0,1)$, see the comment to this question by @Did here. Unfortunately I am not even sure on how to (tightly) lower bound the right integral for this special case.
Any help on solving the general case is much appreciated.
Let $M_n:=\max_{1\le i\le n}|X_i|$. Assume w.l.o.g. that $\sigma=1$. Then, using Markov's inequality ($n\ge 2$), for any $c>0$, \begin{align} \mathsf{E}M_n^2&\ge c\ln n\times\mathsf{P}(M_n\ge \sqrt{c\ln n}) \\[0.5em] &=c\ln n\times (1-[\mathsf{P}(|X_1|<\sqrt{c\ln n})]^n). \end{align}
Using the bound on the error function: $\operatorname{erf}(x)\le \sqrt{1-\exp(-4x^2/\pi)}$, one gets \begin{align} \mathsf{P}(|X_1|< \sqrt{c\ln n})&=\operatorname{erf}\left(\sqrt{c\ln(n)/2}\right) \\ &\le \sqrt{1-\exp(-2c\ln(n)/\pi)} \\ &=\sqrt{1-n^{-2c/\pi}}. \end{align}
Thus, for $c\in (0,\pi/2)$, $$ \mathsf{E}M_n^2\ge c\ln n\times\left(1-\left(1-n^{-2c/\pi}\right)^{n/2}\right)=c\ln n\times(1-o(1)). $$