Surface area of superellipsoid (dice)
You can't expect a closed formula for this surface area. The perimeter of an ellipse, much less the perimeter of a superellipse or the surface area of an ellipsoid or a superellipsoid, is already an integral that doesn't have a formula in the usual sense of an elementary formula. Instead, people did what they always do when an integral is meaningful but has no formula: They named it. That one is called an elliptic integral, but it's just for the case $m=2$. For the $m=2$ case, there is a way out in 3 dimensions: After you define elliptic functions to be the answer in 2 dimensions (for not just the full perimeter but the perimeter of arcs), you can express the 3-dimensional surface area in terms of them without naming any more functions. However, I don't expect that to happen for general $m$, which in any case only reduces the dimension of the problem by 1 instead of solving it.
It's true that your superellipsoids are in the special case that the three semiradii are equal, but I don't think that that rescues you for general values of $m$. Obviously it does rescue you when $m=2$.
What you really want to do for your problem is integrate numerically. You might want a convenient initial change of variables such as $$(a,b,c) = (|x/r|^m,|y/r|^m,|z/r|^m),$$ so that you then get an integral over the triangle $a+b+c=1$. After that there are a variety of numerical integration methods that converge quickly. For instance, you can do the double integral by doing Gaussian integration twice. This particular strategy is not robust as $m \to \infty$, but you didn't say that you needed that limit. An alternative which I guess is robust in that limit, in the case that the semiradii are all equal, is to integrate over one 1/24 region of the spherical angle with the change of coordinates: $$(x,y,z) \propto (1,u,v)$$ with $0 \le u \le v \le 1$. That is also an integral over a triangle.