The error is that you assume that the density distribution is "nearly spherically symmetric". It's far enough from spherical symmetry if you want to calculate first-order subleading effects such as the equatorial bulge. If your goal is to compute the deviations of the sea level away from the spherical symmetry (to the first order), it is inconsistent to neglect equally large, first-order corrections to the spherical symmetry on the other side - the source of gravity. In other words, the term $hg$ in your potential is wrong.
Just imagine that the Earth is an ellipsoid with an equatorial bulge, it's not spinning, and there's no water on the surface. What would be the potential on the surface or the potential at a fixed distance from the center of the ellipsoid? You have de facto assumed that in this case, it would be $-GM/R+h(\theta)g$ where $R$ is the fixed Earth's radius (of a spherical matter distribution) and $R+h(\theta)$ is the actual distance of the probe from the origin (center of Earth). However, by this Ansatz, you have only acknowledged the variable distance of the probe from a spherically symmetric source of gravity: you have still neglected the bulge's contribution to the non-sphericity of the gravitational field.
If you include the non-spherically-symmetric correction to the gravitational field of the Earth, $hg$ will approximately change to $hg-hg/2=hg/2$, and correspondingly, the required bulge $\Delta h$ will have to be doubled to compensate for the rotational potential. A heuristic explanation of the factor of $1/2$ is that the true potential above an ellipsoid depends on "something in between" the distance from the center of mass and the distance from the surface. In other words, a "constant potential surface" around an ellipsoidal source of matter is "exactly in between" the actual surface of the ellipsoid and the spherical $R={\rm const}$ surface.
I will try to add more accurate formulae for the gravitational field of the ellipsoid in an updated version of this answer.
Update: gravitational field of an ellipsoid
I have numerically verified that the gravitational field of the ellipsoid has exactly the halving effect I sketched above, using a Monte Carlo Mathematica code - to avoid double integrals which might be calculable analytically but I just found it annoying so far.
I took millions of random points inside a prolate ellipsoid with "radii" $(r_x,r_y,r_z)=(0.9,0.9,1.0)$; note that the difference between the two radii is $0.1$. The average value of $1/r$, the inverse distance between the random point of the ellipsoid and a chosen point above the ellipsoid, is $0.05=0.1/2$ smaller if the chosen point is above the equator than if it is above a pole, assuming that the distance from the origin is the same for both chosen points.
Code:
{xt, yt, zt} = {1.1, 0, 0};
runs = 200000;
totalRinverse = 0;
total = 0;
For[i = 1, i < runs, i++,
x = RandomReal[]*2 - 1;
y = RandomReal[]*2 - 1;
z = RandomReal[]*2 - 1;
inside = x^2/0.81 + y^2/0.81 + z^2 < 1;
total = If[inside, total + 1, total];
totalRinverse =
totalRinverse +
If[inside, 1/Sqrt[(x - xt)^2 + (y - yt)^2 + (z - zt)^2], 0];
]
res1 = N[total/runs / (4 Pi/3/8)]
res2 = N[totalRinverse/runs / (4 Pi/3/8)]
res2/res1
Description
Use the Mathematica code above: its goal is to calculate a single purely numerical constant because the proportionality of the non-sphericity of the gravitational field to the bulge; mass; Newton's constant is self-evident. The final number that is printed by the code is the average value of $1/r$. If {1.1, 0, 0} is chosen instead of {0, 0, 1.1} at the beginning, the program generates 0.89 instead of 0.94. That proves that the gravitational potential of the ellipsoid behaves as $-GM/R - hg/2$ at distance $R$ from the origin where $h$ is the local height of the surface relatively to the idealized spherical surface.
In the code above, I chose the ellipsoid with radii (0.9, 0.9, 1) which is a prolate spheroid (long, stick-like), unlike the Earth which is close to an oblate spheroid (flat, disk-like). So don't be confused by some signs - they work out OK.
Bonus from Isaac
Mariano C. has pointed out the following solution by a rather well-known author:
http://books.google.com/books?id=ySYULc7VEwsC&lpg=PP1&dq=principia%20mathematica&pg=PA424#v=onepage&q&f=false
Best Answer
This theorem is Proposition LXXI, Theorem XXXI in the Principia. To warm up, consider the more straightforward proof of the preceding theorem, that there's no inverse-square force inside of a spherical shell:
The crux of the argument is that the triangles HPI and LPK are similar. The mass enclosed in the small-but-near patch of sphere HI goes like the square of the distance HP, while the mass enclosed in the large-but-far patch of sphere KL, with the same solid angle, goes like the square of the distance KP. This mass ratio cancels out the distance-squared ratio governing the strength of the force, and so the net force from those two patches vanishes.
For a point mass outside a shell, Newton's approach is essentially the same as the modern approach:
One integral is removed because we're considering a thin spherical shell rather than a solid sphere. The second integral, "as the semi-circle AKB revolves about the diameter AB," trivially turns Newton's infinitesimal arcs HI and KL into annuli.
The third integral is over all the annuli in the sphere, over $0\leq\phi\leq\tau/2$ or over $R-r\leq s\leq R+r$. This one is a little bit hairy, even with the advantage of modern notation.
Newton's clever trick is to consider the relationship between the force due to the smaller, nearer annulus HI and the larger, farther annulus KL defined by the same viewing angle (in modern notation, $d\theta$). If I understand correctly he argues again, based on lots of similar triangles with infinitesimal angles, that the smaller-but-nearer annulus and the larger-but-farther annulus exert the same force at P. Furthermore, he shows that the force doesn't depend on the distance PF, and thus doesn't depend on the radius of the sphere; the only parameter left is the distance PS (squared) between the particle and the sphere's center. Since the argument doesn't depend on the angle HPS, it's true for all the annuli, and the theorem is proved.