You can start by uniformly selecting points on the (n − 1)-sphere and then ask what is the infinity norm of such points.

Given a vector $Z$ with $n$ coordinates taken i.i.d. from the normal distribution $Z_i \sim \mathcal{N}(0, 1)$. The CDF of the absolute maximum of those is given by considering the CDF of $n$ points drawn from the half normal distribution  $F_{||Z||_{\infty}}(x)={\operatorname{erf}\left( \frac{ x }{\sqrt 2} \right) }^{n}$. Now we need to normalize the points, so they'll be on the unit sphere. Given that the positive squared root of the sum of squared normal variables has a Chi distribution with $n$ degress of freedom, $||Z||_2\sim \chi_n$, we have:

$$P(||X||_{\infty} < r\;|\;||X||_2=1) = \\P(||Z||_{\infty}/||Z||_2 < r) = \\\int_{0}^{\infty} P(||Z||_{\infty}/ ||Z||_2 < r\; |\; ||Z||_2 = x)  f_{\chi_n}(x)\;dx = \\ 
\\\int_{0}^{\infty} F_{||Z||_{\infty}}(x \cdot r) f_{\chi_n}(x)\;dx = \\\int_{0}^{\infty} {\operatorname{erf}\left( \frac{ x \cdot r}{\sqrt 2} \right) }^{n}  \cdot \dfrac{x^{n-1}e^{-x^2/2}}{2^{n/2-1}\Gamma\left(\frac{n}{2}\right)}\; dx$$

**Update**

Alternatively, *I'm not entirely sure we can do this*, but I think we can just consider the expected scale of $r$, instead of calculating it as above. It is the mean of $\chi_n$, so the above should be the same as:

$$
F_{||Z||_{\infty}}(\mu(\chi_n) \cdot r)
$$

**Update2**

These formulations seem to be correct for large enough $n$, but not in general. I think this is because $Z_i/||Z||_2$ in not a precise formulation of *the distribution* on the sphere:

[![enter image description here][1]][1]
[![enter image description here][2]][2]


  [1]: https://i.sstatic.net/IRi6R.png
  [2]: https://i.sstatic.net/6I9s5.png