Why is 2s more stable than 2p, if 2p experiences greater effective nuclear charge?
One big issue here is that we generally don't talk about different ionization energies of anything but the outermost orbitals. Once electrons are in "core," we usually don't care since they generally do not participate in any kind of everyday chemistry.

We generally consider the $2s$ orbital to be lower in energy than the $2p$ because the $2s$ has some density inside the radial node that we think of as "nuclear penetration," i.e., the $2s$ electrons are able to get closer to the nucleus through the $1s$ core electrons. Now, once you add a $3s$ electron, I wonder if the innermost radial probability regions of the $3s$ actually repel the $2s$ electrons more than the $2p$ electrons which are generally farther away. The differential repulsion would have to happen in such a way so that it more than compensates for any nuclear penetration effects of the $2s$ electrons. This would certainly align with the observation that the $2p$ electrons are more stabilized only when we add a $3s$ electron.