Solution:
Answer: $\frac{\sqrt{2}}{3}$

Observe that $P$ is a truncated octahedron, formed by cutting off the corners from a regular octahedron with edge length $3$. So, to compute the value of $P$, we can find the volume of the octahedron, and then subtract off the volume of truncated corners.

Given a square pyramid where each triangular face is an equilateral triangle, and whose side length is $s$, the height of the pyramid is $\frac{\sqrt{2}}{2} s$, and thus the volume is $\frac{1}{3} \cdot s^{2} \cdot \frac{\sqrt{2}}{2} s = \frac{\sqrt{2}}{6} s^{3}$.

The side length of the octahedron is $3$, and noting that the octahedron is made up of two square pyramids, its volume must be $2 \cdot \frac{\sqrt{2}(3)^{3}}{6} = 9 \sqrt{2}$.

The six "corners" that we remove are all square pyramids, each with volume $\frac{\sqrt{2}}{6}$, and so the resulting polyhedron $P$ has volume $9 \sqrt{2} - 6 \cdot \frac{\sqrt{2}}{6} = 8 \sqrt{2}$.

Finally, to find the volume of all points closer to one particular vertex than any other vertex, note that due to symmetry, every point in $P$ (except for a set with zero volume), is closest to one of the $24$ vertices. Due to symmetry, it doesn't matter which $V$ is picked, so we can just divide the volume of $P$ by $24$ to obtain the answer $\frac{\sqrt{2}}{3}$.