Let $Z_i:=x_i$ and $M:=M_n:=\|x\|_\infty=\max_1^n|Z_i|$.
By rescaling, without loss of generality $\sigma=1$. So, for real $u>0$
\begin{multline}
P(M^2>u)=P(M>\sqrt u)=1-P(\max_1^n|Z_i|\le\sqrt u)=1-P(|Z_1|\le\sqrt u)^n \\
=1-(1-2G(\sqrt u))^n=1-e^{-ng(u)}, \tag{1}
\end{multline}
where
$$G(x):=P(Z_1>x)\sim\frac1{x\sqrt{2\pi}}e^{-x^2/2}
$$
as $x\to\infty$ and
$$g(u):=-\ln(1-2G(\sqrt u))\sim2G(\sqrt u)\sim\frac2{\sqrt{2\pi u}}e^{-u/2}
=e^{-u/(2+o(1))}
$$
as $u\to\infty$.
Also, $g(u)$ decreases from $\infty$ to $0$ as $u$ increases from $0$ to $\infty$. So, for each natural $n\ge3$ there are unique positive real numbers $u_n$ and $v_n$ such that
$$ng(u_n)=\ln n,\quad ng(v_n)=1.
$$
Clearly, $0<u_n<v_n<\infty$. Also,
$$\frac{\ln n}n=g(u_n)=e^{-u_n/(2+o(1))}\quad\text{and}\quad
\frac{\ln n}n=e^{-(1+o(1))\ln n},
$$
whence
$$u_n\sim2\ln n\quad\text{and, similarly,}\quad v_n\sim2\ln n.
$$
Next,
\begin{equation}
EM^2=\int_0^\infty P(M^2>u)\,du=\int_0^\infty (1-e^{-ng(u)})\,du=I_1+I_2+I_3, \tag{2}
\end{equation}
where
$$I_1:=\int_0^{u_n}(1-e^{-ng(u)})\,du,\quad I_2:=\int_{u_n}^{v_n}(1-e^{-ng(u)})\,du,\quad I_3:=\int_{v_n}^\infty (1-e^{-ng(u)})\,du.
$$
If $0<u<u_n$, then $0<e^{-ng(u)}<e^{-ng(u_n)}=1/n$. So,
$$I_1\sim u_n.$$
Next,
$$I_2\le v_n-u_n=o(u_n),
$$
$$I_3<\int_{v_n}^\infty ng(u)\,du\sim
\int_{v_n}^\infty n\frac2{\sqrt{2\pi u}}e^{-u/2}\,du
\sim 2n\frac2{\sqrt{2\pi v_n}}e^{-v_n/2}
\sim 2ng(v_n)=2=o(u_n).
$$
We conclude that, for $\sigma=1$,
$$E\|x\|_\infty^2=EM^2\sim u_n\sim2\ln n.
$$
So, for any real $\sigma>0$,
$$E\|x\|_\infty^2\sim2\sigma^2\ln n.
$$
Along the same lines, one can give an explicit non-asymptotic lower bound on $E\|x\|_\infty^2$ which will be asymptotically equivalent to $E\|x\|_\infty^2$ as $n\to\infty$. Indeed, one can use the inequality
$$G(t)\ge B(t):=\frac{f(t)}{\sqrt{t^2+2}}
$$
for real $t\ge0$, where $f$ is the standard normal pdf, so that $f(t)=\frac1{\sqrt{2\pi}}\,e^{-t^2/2}$ for real $t$. The latter lower bound on $G(t)$ is a simpler, even if a bit less accurate, version of Birnbaum's lower bound $B_1(t):=f(t)(\sqrt{t^2+4}-t)/2$ on $G(t)$; we have $B_1(t)>B(t)$ for all real $t\ge0$.
So, assuming that $\sigma=1$ and $n\ge16$, letting
\begin{equation*}
w_n:=2\ln n-2\ln\ln n,
\end{equation*}
and recalling (1), for $u\in[0,w_n]$ we have $\ln\ln n>1$ and
\begin{align*}
P(M^2>u)&=1-(1-2G(\sqrt u))^n \\
&\ge1-\exp\big\{-2nG(\sqrt u)\big\} \\
&\ge1-\exp\big\{-2nG(\sqrt w_n)\big\} \\
&\ge1-\exp\Big\{-2n\frac{f(\sqrt w_n)}{\sqrt{w_n+2}}\Big\} \\
&=1-\exp\Big\{-\frac1{\sqrt\pi}\frac{\ln n}{\sqrt{1+\ln n-\ln\ln n}}\Big\} \\
&\ge1-\delta_n,
\end{align*}
where
\begin{equation*}
\delta_n:=\exp\Big\{-\sqrt{\frac{\ln n}\pi}\,\Big\}\to0.
\end{equation*}
Now it follows from (2) that, for $\sigma=1$,
\begin{equation*}
E\|x\|_\infty^2=EM^2\ge\int_0^{w_n} P(M^2>u)\,du
\ge(1-\delta_n)w_n=(1-\delta_n)(1-\epsilon_n)2\ln n,
\end{equation*}
where
\begin{equation*}
\epsilon_n:=\frac{\ln\ln n}{\ln n}\to0.
\end{equation*}
So, for any real $\sigma>0$ and any $n\ge16$,
$$E\|x\|_\infty^2\ge2(1-\delta_n)(1-\epsilon_n)\sigma^2\ln n\sim2\sigma^2\ln n,
$$
as claimed.
For $n=1,\dots,15$, the values of $E\|x\|_\infty^2$ can be easily computed numerically, using (2) and (1), with any degree of accuracy, say to get these $15$ approximate values for $E\|x\|_\infty^2/\sigma^2$: $1, 1.63662, 2.10266, 2.47021, 2.77375, 3.03236, 3.25771, 3.45743,
3.6368, 3.79962, 3.9487, 4.08621, 4.21382, 4.33288, 4.44447$.
