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 $u>0$ 
$$P(M^2>u)=P(M>\sqrt u)=1-P(\max_1^n|Z_i|\le\sqrt u)=1-P(|Z_1|\le\sqrt u)^n
=1-(1-2G(\sqrt u))^n=1-e^{-ng(u)}, 
$$
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,
$$EM^2=\int_0^\infty P(M^2>u)\,du=\int_0^\infty (1-e^{-ng(u)})\,du=I_1+I_2+I_3,
$$
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 
$$E\|x\|_\infty^2=EM^2\sim u_n\sim2\ln n.
$$
So, for any real $\sigma>0$,
$$E\|x\|_\infty^2\sim2\sigma^2\ln n.
$$