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
The Wikipedia article you linked states:
Atomic clocks show that a modern day is longer by about 1.7 milliseconds than a century ago
If we take this change of 1.7 ms/century and multiply by 2.5 million centuries (250 million years) then we get a change of 4,250 seconds or 1.18 hours. So 250 million years ago the day length would have been 22.82 hours.
The circumference of the Earth around the equator is 40,075 km, so the speed of rotation at the equator would have been about 1,750 km/hr or about 1,092 mph. The current speed is 1,670 km/hr or about 1,040 mph.
Later:
If you're interested, the paper "Geological constraints on the Precambrian history of Earth's rotation and the Moon's orbit", Reviews of Geophysics 38 (1): 37–60, 2000, by George E. Williams discusses the day length changes since the Precambrian. There is a PDF available here. From his figure 2 the estimate of 22.8 hours 250 million years ago looks pretty close.
Best Answer
How fast would a sphere need to rotate for a dust speck at its equator to achieve balance between gravitational attraction and centrifugal force?
If you do the math (equating $G M m / R^2$ to $m \omega^2 R$ and using $M = \frac{4\pi}{3} \rho R^3$ as well as $\omega = 2\pi f$), it follows that the size of the sphere is entirely irrelevant and that only the density $\rho$ of the sphere enters into the equation for $f$, the number of revolutions per unit time: $$f^2 = \frac{1}{3\pi}G\rho$$ For $\rho = 5.5 \times 10^3$ kilogram per cubic meter (the density of planet earth) it follows that $f=0.197 \times 10^{-3}$ revolutions per second, corresponding to a revolution period of $5070$ seconds (1 hour and 24 minutes).