Opposite edges are equal in pairs, so the tetrahedron fits in a rectangular box $a\times b\times c$ with its edges as face diagonals of the box.
$a^2+b^2=25$, $a^2+c^2=34$, $b^2+c^2=41$ give $a^2=9,\ b^2=16,\ c^2=25$, so the box is $3\times4\times5$.
The box is cut by the tetrahedron into the tetrahedron and four corner tetrahedra, each with three mutually perpendicular edges $3,4,5$; each has volume $\frac16\cdot3\cdot4\cdot5=10$. Hence $V=60-4\cdot10=\boxed{20}$.
