for N>3, the intersections of x1+x2+...+xN=N/2 with the axes are no longer located inside the cube. bummer

Following 插老 and 网警's suggestion, I got the fomula to calculate the volume of the corner of the cube.

x1+x2+...+xN=a cut a corner from a cube in N-D space. The intersection is a simplex of N-1 D. The height h = a /sqrt(N), the volume is V= a^N / N!

For N=1,2,3, let a=0.5N-0.5, V is the result of JFF's problem.

For N>3, a>1, which means the corner is no longer confined inside JFF's cube.