5. $\frac{\sqrt{66}}{2}$.

From $A B^{2}+B C^{2}=1^{2}+(2 \sqrt{6})^{2}=25=A C^{2}$,
$$
B C^{2}+C D^{2}=(2 \sqrt{6})^{2}+5^{2}=49=B D^{2} \text {, }
$$

we know $A B \perp B C, B C \perp C D$.
As shown in Figure 2, construct rectangle $A B C E$.
Thus, $V_{\text {pyramid } D-A B C}$
From $A E \perp E C$,
$$
A E \perp C D,
$$

we know $A E \perp$ plane $E C D$.
Also, from $A E \perp E D$ and

the Pythagorean theorem, we get $D E=5$.
Furthermore, since the sides of $\triangle E C D$ are $1, 5, 5$, its area is $\frac{3 \sqrt{11}}{4}$.
Thus, $V_{\text {pyramid } A-E C D}=\frac{1}{3} \times 2 \sqrt{6} \times \frac{3 \sqrt{11}}{4}=\frac{\sqrt{66}}{2}$
$$
\Rightarrow V_{\text {tetrahedron } A B C D}=\frac{\sqrt{66}}{2} .
$$
5. In the tetrahedron $ABCD$,
\[
\begin{array}{l}
AB=1, BC=2 \sqrt{6}, CD=5, \\
DA=7, AC=5, BD=7 .
\end{array}
\]

Then its volume is