5. Given that the side length of a regular tetrahedron is 1, and the dihedral angle between a side face and the base is $45^{\circ}$. Then the volume of the circumscribed sphere of the regular tetrahedron is $\qquad$ .
5. $\frac{5 \sqrt{5} \pi}{6}$.

Let the regular tetrahedron be denoted as $P-ABC$, with $H$ being the projection of point $P$ on the base, and $M$ being the midpoint of edge $AB$. Then $H$ is the center of the equilateral triangle $\triangle ABC$, and $HM \perp AB$.
From the given conditions, we know that $\angle PMH=45^{\circ}$.
Let $HM=HP=a$, and $PM=\sqrt{2}a$. Then
$$
\begin{array}{l}
HA=2a \\
\Rightarrow PA=\sqrt{HA^{2}+HP^{2}}=\sqrt{5}a=1 \\
\Rightarrow a=\frac{\sqrt{5}}{5} .
\end{array}
$$

Let $O$ be the center of the circumscribed sphere of the regular tetrahedron $P-ABC$.
By symmetry, point $O$ lies on the ray $PH$, and let the radius of the circumscribed sphere $PO=r$, then
$$
\begin{array}{l}
r^{2}=AO^{2}=HA^{2}+(PO-PH)^{2} \\
=(2a)^{2}+(r-a)^{2}=r^{2}+5a^{2}-2ar \\
\Rightarrow r=\frac{5a}{2}=\frac{\sqrt{5}}{2} .
\end{array}
$$

Thus, the volume of the circumscribed sphere of the regular tetrahedron is
$$
\frac{4}{3} \pi r^{3}=\frac{5 \sqrt{5} \pi}{6} .
$$