jmc

Let $A,B,C,$ and $D$ be the corners of a regular tetrahedron of side length $1$. Let $P$ be the foot of the perpendicular from $D$ to face $ABC$, and let $h$ be the height $DP$: Then, by the Pythagorean theorem, we have $$h^2+(PA{)}^2=h^2+(PB{)}^2=h^2+(PC{)}^2=1,$$so $PA=PB=PC$. The only point on face $ABC$ that is equidistant from $A,B,$ and $C$ is the intersection of the altitudes. If $M$ is the midpoint of $AC$, then $△CPM$ is a $30^{\circ }$-$60^{\circ }$-$90^{\circ }$ triangle with $CM=\frac{1}{2}$, so $PC=\frac{2}{\sqrt{3}}\cdot \frac{1}{2}=\frac{1}{\sqrt{3}}$.

Therefore, $$h=\sqrt{1-(PC{)}^2}=\sqrt{1-{\left(\frac{1}{\sqrt{3}}\right)}^2}=\sqrt{1-\frac{1}{3}}=\sqrt{\frac{2}{3}}=\frac{\sqrt{2}}{\sqrt{3}},$$and the volume of tetrahedron $ABCD$ is $$\begin{array}{rl}V& =\frac{1}{3}\cdot (\text{area of }△ABC)\cdot h\\ & =\frac{1}{3}\cdot \left(\frac{1}{2}\cdot 1\cdot \frac{\sqrt{3}}{2}\right)\cdot \frac{\sqrt{2}}{\sqrt{3}}\\ & =\frac{\sqrt{2}}{12};\end{array}$$the square of the volume is $$V^2={\left(\frac{\sqrt{2}}{12}\right)}^2=\frac{2}{144}=\boxed{\frac{1}{72}}.$$