In $2$ dimensions, a $2$-sphere can be divided into $2^2 = 4$ congruent pieces, the $4$ quadrants, each of angle $\pi/2$ radians.
In $3$ dimensions, a $3$-sphere can be divided into $2^3 = 8$ congruent pieces, the $8$ octants, each of angle $(\pi/2)^2$ radians.
But what about in higher dimensions?
Do we have? : In $n$ dimensions, a $n$-sphere can be divided into $2^n$ congruent pieces, the $n$ hyper-octants, each of angle $\pi/2$.
It would be logical... But the surface of a $4$-sphere in $4$ dimensions is not equal to $2^4 \pi/2 = 8 \pi$ cube radian. It's equal to $2 \pi^2$ cube radian.
So apparently, in $4$-dimensions, each of the $2^{4} = 16$ hyper-octant has an angle of $\dfrac{(\pi/2)^2}{2} = \dfrac{\pi^2}{8}$ cube radian.
The surface of a $n$-sphere in $n$ dimensions is equal to: $\dfrac{2^n (\pi/2)^{\lfloor n/2\rfloor}}{n!!}$
Therefore, it would seem that in $n$-dimensions, each of the $2^n$ hyper-octant of the $n$-sphere has an angle of $\dfrac{(\pi/2)^{\lfloor n/2\rfloor}}{n!!}$.
So does that mean that, in $n>3$ dimensions, the $2^n$ hyper-octants of a $n$-sphere don't have a $n$-dimensional right angle?
Would that mean that a right angle ($\pi/2$) is something very fundamental in $2$ and $3$ dimensions, but it's unimportant and meaningless in any higher dimension?
It is true that the $(n-1)$-dimensional unit sphere $S^{n-1}\subset{\mathbb R}^n$ can be divided into $2^n$ congruent hyperoctants. The "spatial angle" enclosed by such a hyperoctant is $$=2^{-n}\omega_{n-1}(S^{n-1})\ ,$$ where $\omega_{n-1}$ denotes $(n-1)$-dimensional surface area. Envisaging $S^{n-1}$ as an "infinitesimal shell" of thickness $\epsilon\to0$ gives the formula $$\omega_{n-1}(S^{n-1})=\lim_{\epsilon\to0+}{{\rm vol}(B_{1+\epsilon}\setminus B_1)\over\epsilon}={d\over dr}{\rm vol}(B_r)\biggr|_{r=1}=n\>\kappa_n\ ,$$ where $\kappa_n$ denotes the $n$-dimensional volume of the unit ball $B_1\subset{\mathbb R}^n$. It is an exercise in multivariable calculus to obtain the recursion formula $$\kappa_n={2\pi\over n}\>\kappa_{n-2}\qquad(n\geq3)\ .$$ Since $\kappa_1=2$ and $\kappa_2=\pi$ we get $\kappa_3={4\pi\over3}$ (as expected) and then $\kappa_4={\pi^2\over2}$.
It follows that when $n=4$, the $2^4=16$ hyperoctants on the unit sphere $S^3$ each have an area (= "the enclosed spatial angle") given by $${1\over 16}\omega_3(S^3)={1\over16}\cdot 4\>\kappa_4={\pi^2\over8}\ ;$$ and for $n>4$ you'll obtain other such constants involving powers of $\pi$ and combinatorial factors.
Now these values are just certain positive numbers that can be expressed in terms of $\pi$. They have nothing to to with "radians" and right angles. Note that an angle is something "linear" measured in two dimensions, and making sense only in this realm. Given any two linearly independent vectors ${\bf x}$, ${\bf y}\in{\mathbb R}^n$ they span a two-dimensional plane, and their enclosed angle makes sense as a usual planar angle. It computes to $$\angle({\bf x},{\bf y})=\arccos{{\bf x}\cdot{\bf y}\over|{\bf x}|\ |{\bf y}|}\ .$$ Contrasting this the numbers we obtained above were $(n-1)$-dimensional surface measures.