Does earth rotate in a shorter time at 0m than at 5000m? We know rotation period of earth (stellar day) is 86164.098 903 691 seconds of mean solar time.
At 0m time is supposed to be more dilated than at 5000m because of gravitational time dilation.
Then hypothetically we count less nanoseconds at 0m relative to 5000m of altitude.
Then assuming this is true, stellar day is shorter at 0m than at 5000m.
Doesn't it mean we have a perception of a quicker rotation of earth at 0m than at 5000m? A quicker rotation in a dilated frame? Does it make sense?

If you agree that rotation period of earth (stellar day) is the same at all altitudes, why isn't it affected by time dilation?
 A: 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.
A: 
Does earth rotate in a shorter time at 0m than at 5000m?

No. If it did, it would be twisting, and it isn't. The Earth's rate of rotation relative to the fixed stars is the same at all elevations. It takes a day for a point at sea level to go round full circle, and it takes the same time for a point at 5000m to go round full circle. 

We know rotation period of earth (stellar day) is 86164.098 903 691 seconds of mean solar time. At 0m time is supposed to be more dilated than at 5000m because of gravitational time dilation.

It is. Clocks go slower when they're lower.  

Then hypothetically we count less nanoseconds at 0m relative to 5000m of altitude.

Not just hypothetically. See the interview with David Wineland of NIST where he's talking about optical clocks: "If one clock in one lab is 30 centimeters  higher than the clock in the other lab, we can see the difference in the rates they run at".   

Then assuming this is true, stellar day is shorter at 0m than at 5000m.

Yes, though it's a technicality. Your clocks run slower when you're lower, so your measurement is different. Meanwhile the Earth turns in its own sweet time at the same rate at all elevations.  

Doesn't it mean we have a perception of a quicker rotation of earth at 0m than at 5000m? A quicker rotation in a dilated frame?

Yes it does. But it's our perception. The Earth isn't really rotating quicker at 0m. Besides, the difference we perceive is very slight. Gravitational time dilation at the surface of the Earth is only about one part in 10-9. We hardly notice it, though we do need to take account of it in our GPS satellites. It's much more significant for a neutron star. A clock at the surface of a neutron star would be going at circa 0.8 times the rate of a clock at the surface of the Earth.  
