I think you can start with the process of uniformly selecting a point on (n − 1)-sphere and then consider the original samples as independent.
So we draw a vector $Y$ with $n$ coordinates taken i.i.d. from the normal distribution $Y_i \sim \mathcal{N}(0, 1)$. The CDF of the absolute maximum of those is given by taking the maximum of $n$ points from the half normal distribution $F_{||Y||_{\infty}}(x)={\operatorname{erf}\left( \frac{ x }{\sqrt 2} \right) }^{n}$. We do need to normalize the points, so they'll be on the unit sphere. Given that the sum of squares of normal variables is Chi-squared, $||Y||_2^2\sim \chi^2(n)$, we have:
$$P(||X||_{\infty} < r) = P(||Y||_{\infty}/|Y||_2 < r) = \\\int_{0}^{\infty} P(||Y||_{\infty}/\sqrt{x} < r\; |\; ||Y||_2^2 = x) f_{\chi^2}(x; n)\;dx$$
so the closed form solution:
$$\int_{0}^{\infty} {\operatorname{erf}\left( \frac{ \sqrt x \cdot r}{\sqrt 2} \right) }^{n} f_{\chi^2}(x; n)\; dx$$