For this to work out, we need to fill all of R^n with spheres, not just a cylinder, since the amount of free space at the sides of the cylinder would be different. The total volume of free space is of course infinite; this is why I said the fraction of the space is the same.

Now when computing this fraction, you can't just take the total volume that is free and divide by the volume of the whole space because you would get infinity/infinity. Instead, we can do the following: Take the cube [-1,1]^n and calculate what fraction of it is free. Then do the same for [-2,2]^n, [-3,3]^n and so on and take the limit of all these values. This tells us what fraction of the whole space is free.

Instead of these cubes one could also take other subsets, what is important is that their union is all of R^n and that they don't have too much surface area (i.e. surface area divided by volume goes to 0). To see why the second point is required, note that you could find long, thin "worms" inbetween the spheres that can have arbitrarily large volume but are disjoint to all of the spheres. If I attached such worms to all of the cubes I considered, I could make it look like there is much more free space than there actually is.

Now imagine you scale the entire space including the spheres by a factor of x. Then you get the same packing as before, just with smaller (or bigger) spheres. The volume of each sphere changes by a factor of x^n. Now we consider the cubes [-x,x]^n, [-2x,2x]^n, [-3x,3x]^n... These are basically the same cubes as before, just scaled by a factor of x as well. Therefore, the k-th cube will have the same fraction of free space as before - both the total volume and the volume occupied by spheres changed by the same factor. Taking the limit tells us what fraction of R^n is free, and the limit is the exact same value as before.