(4) Given a regular tetrahedron $P-ABC$ with the side length of the equilateral triangle base being 1, and the center $O$ of its circumscribed sphere satisfies $\overrightarrow{O A}+\overrightarrow{O B}+\overrightarrow{O C}=\overrightarrow{0}$, then the volume of this regular tetrahedron is $\qquad$.
(4) Since $\overrightarrow{O C}=-(\overrightarrow{O A}+\overrightarrow{O B})$, point $C$ lies on the plane determined by points $O, A, B$, i.e., points $O, A, B, C$ are coplanar, thus point $O$ is on the base plane $A B C$. Also, since $\overrightarrow{O A}+\overrightarrow{O B}+\overrightarrow{O C}=\overrightarrow{0}$, point $O$ is the centroid of $\triangle A B C$. Since $\triangle A B C$ is an equilateral triangle, point $O$ is its center. As shown in the figure, connect $C O, P O$, then
$$
P O=O C=\frac{2}{3} \times \frac{\sqrt{3}}{2} \times 1=\frac{\sqrt{3}}{3},
$$

Therefore, the volume of the regular tetrahedron is
$$
\frac{1}{3} \times S_{\triangle A B C} \times P O=\frac{1}{3} \times \frac{\sqrt{3}}{4} \times 1^{2} \times \frac{\sqrt{3}}{3}=\frac{1}{12} .
$$