OK, I'm sufficiently fed up with this to attempt a mathematical answer. It'll be very approximate, but it'll demonstrate an upper limit to how much the sea is skewed to one side of the planet (it's actually more than I expected). I fully expect to get something wrong here; if it's something that matters, please tell me about it!
Let's imagine a cartoon planet that consists of two parts:
- A sphere of radius 6370 km
- A half-shell of "continentals" with a thickness of 4 km (representing the distance from seafloor to continent)
For simplicity we'll assume that all of these have equal density, and in arbitrary units we'll define that density to be 1.
We want to find where the centre of mass of this "planet" is. Let's define our coordinate system to work in one dimension, along a line running from the middle of one half-shell to the middle of the other, where x=0 is at the centre of the sphere.
Here's a diagram (not to scale):

$r_1=6370$ km; $r_2=6374$ km
The mass of the sphere, in arbitrary mass units, is $\frac{4}{3} \pi r_1^3 = 1.1\times 10^{12}$.
The mass of the continental half-shell is $\frac{1}{2}\cdot\frac{4}{3} \pi (r_2^3 - r_1^3)=10^9$.
The centre of mass (CoM) of the sphere is at $x=0$.
I had to look up online how to find the centre of mass of a hemispherical shell, but it turns out that it lies at $x=\frac{3}{8}\frac{r_2^4 - r_1^4}{r_2^3 - r_1^3} = 3186$ km.
To find the location of the centre of mass of the combined object, we use
$x_c = \frac{m_1d_1 + m_2d_2}{m_1 + m_2}$ where each $m$ is a mass and each $d$ is the position of the centre of mass of that object. So,
$x_c = \frac{1.1 \times 10^{12} \times 0 + 10^9 \times 3186}{1.1\times 10^{12} + 10^9} = 2.9 $ km.
So the centre of mass of our cartoon planet is 2.9 km closer to the "continent" side than the "ocean" side. That means that the ocean will, rather approximately, try to form a shell centred on that point, thus being 2.9 km deeper in the centre of the continent than in the centre of the ocean. Errr...
Except that the sea wouldn't actually do this because it would be blocked by the giant continent that's in its way. To figure out what would actually happen would require doing the maths in 3D, and possibly involving tractive forces and the like, and that's well beyond my capability this afternoon ;-)
I emphasise once again that this is an extremely generous scenario - half of the planet is continent, which is more than on Earth, and everything has the same density, rather than the inside of the planet being denser than the crust (which would make the effect much weaker). Plus, once we're dealing with values of a few km, in the real world there are probably other variations (e.g. equatorial bulge, non-ellipsoisal geoid due to density variations, etc) that have much bigger effects.