Solution:

The three vertices adjacent to $A$ are its two neighbors $B$ and $C$ in the same hexagonal base and the vertex $D$ directly above $A,$ with $DA = h$ perpendicular to the base. The face in the base is $ABC,$ and the face avoiding $A$ is $BCD;$ they meet along $\overline{BC}.$

Let $E$ be the midpoint of $\overline{BC}.$ Since $AB = AC = 12$ and $\angle BAC = 120^\circ$ (the interior angle of a regular hexagon), $\overline{AE} \perp \overline{BC}$ and $AE = 12\cos 60^\circ = 6.$ Because $\overline{DA}$ is perpendicular to the base, $\overline{DE} \perp \overline{BC}$ as well, so the dihedral angle is $\angle DEA = 60^\circ.$

In right triangle $DAE,$ $h = AE\tan 60^\circ = 6\sqrt{3},$ so $h^2 = 108.$