$\newcommand{\si}{\sigma}$
Let $M_n$ be the maximum of $n$ iid standard normal random variables (r.v.'s) $Z_1,\dots,Z_n$, and let $L_N$ be the maximum of $N$ iid standard normal r.v.'s $W_1,\dots,W_N$, where $N:=n(n-1)/2$. Assume also the r.v.'s $M_n$ and $L_N$ are independent. By rescaling, the question can be restated as follows:

Given a positive real $\si$, how does the probability
$P(M_n>\si L_N)$ behave?

The answer to this questions is this:

$$P(M_n>\si L_N)\longrightarrow
\begin{cases}
1&\text{ if }\si<1/\sqrt2, \\
0&\text{ if }\si\ge1/\sqrt2;
\end{cases} \tag{1}
$$
the convergence everywhere here is as $n\to\infty$. Moreover, it does not matter that the r.v.'s $M_n$ and $L_N$ are independent.

Indeed, it is a well-known fact of the so-called extreme value theory (see e.g. Theorem 18.4) that
$$M_n=d_n+\frac{G_n}{\sqrt{2\ln n}}, \tag{2}
$$
where
$$d_n:=\sqrt{2\ln n}-\frac{\ln\ln n+\ln(4\pi)}{2\sqrt{2\ln n}}$$
and $G_n\to G$ in distribution, where in turn $G$ is a Gumbel r.v.

So, $M_n\sim d_n\sim\sqrt{2\ln n}$ and $L_N\sim d_N\sim\sqrt{2\ln N}\sim\sqrt{4\ln n}$ in probability, which immediately implies (1) for $\si\ne1/\sqrt2$.

The case $\si=1/\sqrt2$ (which was of special interest in the original question), a bit more effort shows that
$$\frac1{\sqrt2}\,d(N)-d(n)\sim\frac{\ln\ln n}{4\sqrt{2\ln n}}
>>\frac1{\sqrt{\ln n}},
$$
where $A>>B$ means $B=o(A)$. Now, in view of (2), the case $\si=1/\sqrt2$ follows as well.

Somehow, I have just noticed that the original question was, not about the maximum of $n$ iid standard normal r.v.'s, but about the maximum of **the absolute values of $n$** iid standard normal r.v.'s. Anyway, (1) holds as well with $M_n^{|\;|}:=\max_1^n|Z_i|$ and $L_N^{|\;|}:=\max_1^N|W_j|$ in place of $M_n=\max_1^n Z_i$ and $L_N=\max_1^N W_j$, respectively. Here one only needs to make small adjustments. Indeed, the essential fact used in the above derivation is that
$$M_n=d_n+\frac{\ln\ln n}{\sqrt{\ln n}}\,o_P(1), \tag{3}
$$
where $o_P(1)$ stands for a r.v. depending on $n$ and converging to $0$ in probability. But (3) holds with $M_n^-:=\max_1^n(-Z_i)$ in place of $M_n$, because $M_n^-$ equals $M_n$ in distribution. So, (3) holds with $M_n^{|\;|}$ in place of $M_n$, because every value of $M_n^{|\;|}$ is either the corresponding value of $M_n$ or the corresponding value of $M_n^-$.