$n = 3$ is certainly possible. For example, take $\angle {APB} = \angle {APC} = \angle {BPC} = 90^{o}$ (so that the lines $PA, PB, PC$ are mutually perpendicular). Then the three planes through P are also mutually perpendicular, so the two sums are both $270{o}$. We show that $n > 3$ is not possible. The sum of the n angles $APB$ etc at $P$ is less than $360^o$. This is almost obvious. Take another plane which meets the lines $PA, PB, PC$ etc at $A', B', C', ...$ and so that the foot of the perpendicular from P to the plane lies inside the n-gon $A'B'C' ...$ then as we move P down the perpendicular the angles $A'PB'$ etc all increase. But when it reaches the plane their sum is $360^o$. However, I do not immediately see how to make that rigorous. Instead, take any point O inside the n-gon ABC... . We have ∠PBA + ∠PBC > ∠ABC. Adding the n such equations we get $\sum(180^o - APB) > \sum ABC = (n - 2) 180^o$. So, $\sum APB < 360o$. The sum of the n angles between the planes is at least $(n - 2)*180^o$. If we take a sphere center $P$. Then the lines $PA, PB$ intersect it at $A", B", ...$ which form a spherical polygon. The angles of this polygon are the angles between the planes. We can divide the polygon into $n - 2$ triangles. The angles in a spherical triangle sum to at least $180^o$. So the angles in the spherical polygon are at least $(n - 2) 180^o$. So we have $(n - 2) 180^o < 360^o$ and hence $n < 4$. This solution is copied from the link.Nice solution.