What would be the depth variation of a "water ellipsoid" above a reference ellipsoid Earth?
A complicating factor will be the change in density with temperature. So for this I will assume a constant density of water.
As the thickness of the water layer is quite small compared to the size of Earth (given as 2.6 km by the answer to the other question) I think an approximation with a constant g at any particular location should give a fairly reasonable answer.
Water naturally finds its level based upon pressure. So if we have a constant pressure at the top of the surface, the pressure at the bottom must also be constant. And this should work as a reasonable approximation for any number of layers.
Noting that for any column of water, the pressure increase will be given by:
$P=\rho \cdot g \cdot h$
Taking the values of g from Wikipedia:
At the equator, g, due to Earth being an oblate spheroid and rotating is ~$9.78 \frac{\mathrm{m}}{\mathrm{s}^2}$
At the poles, g is ~$9.83 \frac{\mathrm{m}}{\mathrm{s}^2}$
Equating the 2 we end up with:
$P=\rho \cdot 9.78 \frac{\mathrm{m}}{\mathrm{s}^2} \cdot h_e=\rho \cdot 9.83 \frac{\mathrm{m}}{\mathrm{s}^2} \cdot h_p$
Factoring out common terms leaves us with:
$h_e = \frac{9.83}{9.78} h_p$
So the equator will have water that is roughly 0.5% deeper than the poles.
For 2.6 km, that works out to be roughly 13 m.