proof of uniform hypersphere sampling

Per you comment, if we sample coordinates from a finite ring, the probability distribution function will have to be discreet and so the closest that we can hope for is a discretised normal distribution. A discrete distribution will not be uniform over a continuous space (and also the lengths of the vectors will mean they of not lie on the same sphere unless projected). If $q$ tends to infinity, the variance of the discretised normal is allowed to grow and we normalise lengths, then the distribution will tend to uniform.

We prove the general case of the $n$-sphere. We note that the probability density function of an $n$-fold Gaussian distribution is given by $$\frac1{(2\pi)^{n/2}}\exp(-x_1^2/2)\exp(-x_2^2/2)\cdots\exp(-x_n^2/2)=\frac1{(2\pi)^{n/2}}\exp\left(-(x_1^2+x_2^2+\cdots +x_n^2)\right).$$ Converting to spherical polar coordinates then gives the density function $$\frac{1}{(2\pi)^{n/2}}\exp\left(-r^2\right)$$ which is independent of all of the angular variables and so is uniform on spheres. Rescaling by the radius $r$ therefore gives a variable uniform on the $n$-sphere.

No, the density function in this case is no independent of the angular components. However, by sampling components uniformly you can sample uniformly from the $n$-cube. By either rescaling or selecting a face at random and then selecting a point on the $(n-1)$-cube, you can sample uniformly form the surface of the $n$-cube.