This is a fun calculation to do so let's have a go. What we need to do is calculate the time dilation for an observer rotating with the Earth and see how it changes with height.
To do this we start with the spacetime geometry near the Earth, which is described (approximately) by the Schwarzschild metric:
$$ c^2d\tau^2 = \left(1-\frac{2GM}{c^2r}\right)c^2dt^2 - \frac{dr^2}{1 - \frac{2GM}{c^2r}} - r^2\left(d\theta^2 + \sin^2\theta d\phi^2\right) \tag{1} $$
$\tau$ is the time recorded by a clock carried by our rotating observer, and $t$ is the time recorded by a clock carried by an observer far enough from the Earth for the Earth's gravity to be negligible. The time dilation is then:
$$ \text{time dilation} = \frac{d\tau}{dt} $$
so that's what we are going to calculate.
We'll consider an observer who is stationary at the equator so $\theta=\pi/s$ and $d\theta=0$, and at a distance $r$ from the centre of the Earth. The observer isn't moving radially inwards or outwards so $dr=0$. If we put this lot into equation (1) it simplifies to:
$$ c^2d\tau^2 = \left(1-\frac{2GM}{c^2r}\right)c^2dt^2 - r^2d\phi^2 \tag{2} $$
We need to eliminate $d\phi$ and we do this by noting that if $\omega$ is the angular velocity with which the Earth rotates then:
$$ \frac{d\phi}{dt} = \omega $$
and therefore:
$$ d\phi = \omega dt $$
And we can substitute for $d\phi$ in equation (2) to get:
$$ c^2d\tau^2 = \left(1-\frac{2GM}{c^2r}\right)c^2dt^2 - r^2\omega^2dt^2 $$
And this gives us the equation we want for the time dilation:
$$ \frac{d\tau}{dt} = \sqrt{1-\left(\frac{2GM}{c^2r} + \frac{r^2\omega^2}{c^2}\right)} \tag{3} $$
And equation (3) is the result we need. As we move upwards, i.e. in the direction of increasing $r$, then the term $2GM/c^2r$ decreases and the term $r^2\omega^2/c^2$ increases, and the question is what happens to the sum:
$$ T = \frac{2GM}{c^2r} + \frac{r^2\omega^2}{c^2} $$
If $T$ increases with height (increasing $r$) then time dilation is increasing as we go up while if $T$ decreases with height then time dilation is decreasing as we go up. To find out what happens we just differentiate with respect to $r$:
$$ \frac{dT}{dr} = -\frac{2GM}{c^2r^2} + \frac{2r\omega^2}{c^2} \tag{4} $$
The radius of the Earth at the equator is $r \approx 6378000$m and the angular velocity is $2\pi$ radians in 24 hours so $\omega \approx 7.272 \times 10^{-5}$ radians per second. Put these values into equation (4) and we get:
$$ \frac{dT}{dr} \approx -2.176 \times 10^{-16} + 7.496 \times 10^{-19} \approx -2.169 \times 10^{-16} $$
And there's your answer. At the Earth's surface the time dilation for an observer rotating with the Earth decreases with height i.e. time runs faster as you move upwards. The Earth takes longer to rotate for the higher observer.
