Consider the unit sphere $\mathscr{S}^d$. We can define a hyperspherical cap by the angle $\theta$ that its associated hypercone subtends at the centre of the sphere.
How large does $\theta$ have to be in order that for every set of orthogonal axes, the lines through a set of $d$ mutually orthogonal points on the sphere and the origin, at least one axis intersects on the hyperspherical cap?
This is equivalent to being given a set of axes, and asking for the largest spherical distance away from all of them a point can be. Without loss we can take the standard basis of $\mathbb{R}^d$.
Symmetry considerations imply such a point will be the same distance from all of the axes. Restricting to the section of the sphere with positive coordinates, symmetry also forces it to be the point $d^{-1/2}(1,1,\dotsc,1)$: this is the only point on this side of the sphere an equal distance from the closest points. One then finds the distance $\theta$ satisfies $\cos{\theta} = d^{-1/2}$ from the dot product formula for the distance, and hence $\theta=\operatorname{arcsec}{\sqrt{d}}$.
For example, in two dimensions, $\arccos{(1/\sqrt{2})} = \pi/4$, and in three, $\arccos{(1/\sqrt{3})} \approx 0.955$.