7. The base $ABCD$ of the quadrilateral pyramid $P-ABCD$ is a rhombus with a top angle of $60^{\circ}$, and the angle between each side face and the base is $60^{\circ}$. There is a point $M$ inside the pyramid that is 1 unit away from the base and each side face. Then the volume of the pyramid is
Given $8 \sqrt{3}$.
Let the intersection point of the diagonals $A C, B D$ of the rhombus be $H$. Then $P H$ is both the perpendicular bisector of segment $A C$ and the perpendicular bisector of $B D$, making it the height of the pyramid, and point $M$ lies on $P H$. Thus, the plane $P B D$ is perpendicular to the base plane $A B C D$, and similarly, the plane $P A C$ is also perpendicular to the base plane $A B C D$. The plane $P B D$ divides the pyramid into two equal-volume tetrahedrons, so we only need to consider the tetrahedron $P A B D$.

Let the projection of point $M$ on the plane $P A D$ be $E$, and the plane $M E H$ passes through point $P$ and intersects $A D$ at $F$. Since $\angle M H F=90^{\circ}=\angle M E F$, points $M, E, F, H$ are concyclic. Because $M E \perp$ plane $P A D$, we have $M E \perp A D$. Since $M H \perp$ plane $A B D$, we have $M H \perp A D$. Therefore, $A D \perp$ plane $M E H$, so $A D \perp P F$, and $F H$ is the projection of $P F$ in the plane $A B D$. Thus, $A D \perp F H$, and the dihedral angle $\angle E F H=60^{\circ}$. Therefore, $\angle E M H=120^{\circ}$.

Given $M E=M H=1$, we get $E H=\sqrt{3}$. In the right triangles $M E F$ and $M H F$, $E F=H F$. Since $\angle E F H=60^{\circ}$, $\triangle E F H$ is an equilateral triangle, so $F H=E F=E H=\sqrt{3}$.

In the right triangle $\triangle A F H$, $\angle H A F=30^{\circ}$, so $A H=2 F H=2 \sqrt{3}$. Therefore, the side length of the equilateral $\triangle A B D$ is 4, and thus $S_{\triangle A B D}=4 \sqrt{3}$. In the right triangle $\triangle P F H$, $P H=F H \tan 60^{\circ}=3$, so $V_{P A B D}=\frac{1}{3} P H \cdot S_{A B D}=4 \sqrt{3}$. Therefore, $V_{P-A B C D}=2 V_{P A B D}=8 \sqrt{3}$.