Why is a 3s orbital lower in energy than a 3p orbital in all atoms other than a hydrogen atom which only has a single electron?

Because of how there IS only one electron in `"H"` atom. That single electron does not introduce orbital anguar momentum, so no matter what value of `l`, the orbital energies for the same `n` are all the same.

The added electrons in multi-electron atoms are intrinsically correlated in such a way that each electron is influenced by the motions of the others.

This electron correlation introduces the effect of electron-electron repulsion. That in turn generates an energy splitting of the orbitals with the same `n` but different `l`.

For example, for just the `3s` and `3p`...

`"H"` atom:

`ul(color(white)(uarr darr))" "ul(color(white)(uarr darr))" "ul(color(white)(uarr darr))" "ul(color(white)(uarr darr))"`
`" "" "" "underbrace(" "" "" "" "" "" "" "" "" ")`
`3s" "" "" "" "" "" "3p`

For multi-electron atoms:

`" "" "" "ul(color(white)(uarr darr))" "ul(color(white)(uarr darr))" "ul(color(white)(uarr darr))"`
`" "" "" "underbrace(" "" "" "" "" "" "" "" "" ")`
`ul(color(white)(uarr darr))" "" "" "" "" "3p`
`3s`

We had that the `3s, 3p, 3d` orbitals are the same energy in the `"H"` atom, but in higher-electron atoms, we instead have the energy ordering `3s < 3p < 3d`.

In multi-electron atoms, higher `l` is higher energy for orbitals with the same `n`.