Opposite edges of the tetrahedron have the same length. Let $AB$, $CD$ have length $a$. Take parallel planes through $AB$ and $CD$. Project $A$, $B$ onto the other plane, and $C$, $D$ onto the other plane. Then we get $8$ points at the vertices of a rectangular block. If the sides have lengths $x$, $y$, $z$, then we have $a^2 = y^2 + z^2$, $b^2 = x^2 + z^2$, $c^2 = x^2 + y^2$. A long diagonal of the block has length $\sqrt{x^2 + y^2 + z^2}$ and is a diameter of the circumsphere. So $x^2 + y^2 + z^2 = 4$. Hence $a^2 + b^2 + c^2 = 2(x^2 + y^2 + z^2) = 8$.