Solution:

The displaced volume equals the volume of the part of the cube lying below the plane of the barrel's rim. By symmetry that region is a tetrahedron cut from the bottom corner of the cube: three mutually perpendicular edges of equal length $\ell$ along the cube's edges, capped by an equilateral triangle in the rim plane. The equilateral cross-section is inscribed in the rim circle of radius $4,$ so its side length is $4\sqrt{3},$ and therefore $\ell = \frac{4\sqrt{3}}{\sqrt{2}} = 2\sqrt{6}.$

Taking one of the right isosceles faces as the base, the volume is $$\frac{1}{3}\left(\frac{1}{2}\ell^2\right)\ell = \frac{\ell^3}{6} = \frac{(2\sqrt{6})^3}{6} = \frac{48\sqrt{6}}{6} = 8\sqrt{6}.$$

Thus $v = 8\sqrt{6}$ and $v^2 = 64 \cdot 6 = 384.$