I've come a bit late to this particular party, but here's another argument. This one includes most of the sphere in a suitable cone.
Choose a small positive number x, to be optimized later. Then the volume of that part of the sphere with $0\leq x_n\leq x$ is at most $2^n x$ (the volume of the cube being $2^n$).
Now consider the plane $x_n=x$. This intersects the sphere in an (n-1)-dimensional subsphere. Let C be the smallest cone that contains everything in the sphere that lies above this plane. A simple argument using similar triangles shows that the height of this cone is at most 1/x. Therefore, its volume is at most $2^{n-1}/nx$.
Doubling all this to get both halves of the sphere, we get an upper bond of $2^n(2x+1/nx)$, and taking $x=n^{-1/2}$ we get an upper bound for the ratio of $3n^{-1/2}$.
Of course, this is a weak bound, but I was trying to make the argument as simple as possible. (It's simpler in my head than I've managed to make it written down.)
Apologies if this duplicates someone else's argument -- I checked, but could have missed something.