7. Given a quadrilateral pyramid $P-ABCD$ with the base $ABCD$ being a rhombus with a top angle of $60^{\circ}$, and each side face forms a $60^{\circ}$ angle with the base, a point $M$ inside the pyramid is equidistant from the base and each side face, with the distance being 1. Then the volume of the pyramid is $\qquad$
7. $8 \sqrt{3}$.

Let the intersection point of the diagonals $A C$ and $B D$ of the rhombus be $H$. Then $P H$ is the perpendicular bisector of segments $A C$ and $B D$, i.e., the height of the pyramid $P-A B C D$, and point $M$ lies on $P H$.
Thus, plane $P B D \perp$ plane $A B C D$.
Similarly, plane $P A C \perp$ plane $A B C D$.
Therefore, plane $P B D$ divides the pyramid into two equal-volume tetrahedra.
We only need to consider tetrahedron $P A B D$.
As shown in Figure 2, let the projection of point
$M$ on plane $P A D$ be $E$, and plane $M E H$
pass through point $P$, intersecting $A D$
at point $F$.
Since $\angle M H F=$
$90^{\circ}=\angle M E F$, points
$M, E, F, H$ are concyclic.
By $M E \perp$ plane $P A D$, we have $M E \perp A D$;
By $M H \perp$ plane $A B D$, we have $M H \perp A D$.
Thus, $A D \perp$ plane $M E H$, so $A D \perp P F$.
Since $F H$ is the projection of $P F$ in plane $A B D$, we have $A D \perp F H$, i.e., the dihedral angle $\angle E F H=60^{\circ}$.
Therefore, $\angle E M H=120^{\circ}$.
Given $M E=M H=1$, we have $E H=\sqrt{3}$.
Thus, in right triangles $\triangle M E F$ and $\triangle M H F$, $E F=H F$.
Since $\angle E F H=60^{\circ}$, $\triangle E F H$ is an equilateral triangle.
Therefore, $F H=E F=E H=\sqrt{3}$.
In right triangle $\triangle A F H$, $\angle H A F=30^{\circ}$, so $A H=2 F H=2 \sqrt{3}$.
Thus, the side length of the equilateral $\triangle A B D$ is 4.
Hence, $S_{\triangle A B D}=4 \sqrt{3}$.
In right triangle $\triangle P F H$, $P H=F H \tan 60^{\circ}=3$.
Thus, $V_{\text {tetrahedron } P A B D}=\frac{1}{3} P H \cdot S_{\triangle A B D}=4 \sqrt{3}$.
Therefore, $V_{\text {pyramid } P-A B C D}=2 V_{\text {tetrahedron } P A B D}=8 \sqrt{3}$.