Summary
A perfect sphere will not precess.
Now what about the earth? There are two answers:
- What the video shows is not true precession
- The earth's axis does precess, but that's exactly because the earth is not perfectly spherical
About the first point
You may have noticed that while the earth goes round the sun, the direction of the depicted orbit of the earth does not change. However it is quite obvious from looking on an image of the complete orbit that the direction does change (e.g. if drawn on a vertical plane, on the left of the sun it's vertical, while below the sun it's horizontal). Also, in the video, the sun is always on the right.
So why does the direction not change in the video? Well, it's because the video is in a rotating frame of reference. That is, the "camera" is rotating around the (simulated) earth, in a way that the sun is always on the right of the earth. That's why there's an apparent precession: What you are seeing is the apparent rotation of the complete universe due to the movement of the camera. If background stars had been depicted, you would have seen the stars "rotating" as well.
The best way to see that the earth's axis does not do a full rotation during a single year is to notice that the earth axis does point (approximately) to the polar star both in the summer and in the winter. It wouldn't if the earth's axis would actually process a noticeable amount per year.
Now the video does not show how the earth really rotates today, but how it would rotate in a tidal lock. However the only difference is that the earth would rotate along its axis once a year. Indeed, you'd get the very same movie with our fast-rotating year if you'd make a picture once every 24 hours (well, with some correction for the intervals because of the excentricity of the earth's orbit, but let's ignore those complications).
OK, so how to mathematically describe the movement? Well, if we describe the (active) rotation of the earth with a rotation matrix $R_a(t)$ and the (passive) rotation of the frame of reference with $R_p(t)$, then the rotation is described by the combined rotation $R_p(t)R_a(t)$. Now the passive rotation is in an axis perpendicular to the ecliptic; in the video that's mapped to the $z$ axis. The earth's axis it tilted by an angle $\alpha$ from that direction. So if $R_z(\phi)$ is the rotation matrix around the $z$ axis, $R_y(\phi)$ is the same for the $y$ axis, $\omega_p$ is the angular speed of the earth orbit ($1/\text{year}$) and $\omega_a$ is the angular speed of the earth rotation (in the tidal lock case, $\omega := \omega_a = -\omega_p$), and we put the time $t=0$ at the point when the earth's axis in the rotating frame is tilted in the $y$ direction, the complete rotation matrix for time $t$ for the movie is
$$R(t) = R_z(-\omega t)R_y(-\alpha)R_z(\omega t)R_y(\alpha)$$
Now
$$R_z(\phi) = \begin{pmatrix}
\cos\phi & \sin\phi & 0\\
-\sin\phi & \cos\phi & 0\\
0 & 0 & 1
\end{pmatrix}$$
and
$$R_y(\phi) = \begin{pmatrix}
\cos\phi & 0 & \sin\phi\\
0 & 1 & 0\\
-\sin\phi & 0 & \cos\phi
\end{pmatrix}$$
So the total depicted rotation is described by the rotation matrix
$$
\left(
\begin{array}{ccc}
\cos (\omega t) \sin ^2(\alpha )+\cos (\alpha ) \left(\cos (\alpha ) \cos
^2(\omega t)+\sin ^2(\omega t)\right) & (\cos (\alpha )-1) \cos
(\omega t) \sin (\omega t) & \sin (\alpha ) \left(\sin ^2(\omega
t)+\cos (\alpha ) (\cos (\omega t)-1) \cos (\omega t)\right) \\
\left(\cos (\omega t) \cos ^2(\alpha )-\cos (\omega t) \cos (\alpha
)+\sin ^2(\alpha )\right) \sin (\omega t) & \cos ^2(\omega t)+\cos
(\alpha ) \sin ^2(\omega t) & (\cos (\alpha ) (\cos (\omega t)-1)-\cos
(\omega t)) \sin (\alpha ) \sin (\omega t) \\
\cos (\alpha ) (\cos (\omega t)-1) \sin (\alpha ) & \sin (\alpha ) \sin
(\omega t) & \cos ^2(\alpha )+\cos (\omega t) \sin ^2(\alpha )
\end{array}
\right)
$$
For $\alpha=0$ you'd get the identity matrix. That's because in that case, the passive rotation just cancels out the active rotation.
However note again, that this is not the actual rotation, but the apparent rotation due to the rotating reference frame. For the actual rotation, omit $R_z(-\omega t)$. Then you get a simple rotation around a fixed axis tilted against the $z$ axis by the angle $\alpha$.
About the second point
Now, in reality, the earth's axis does precess, but much more slowly. More exactly, it needs almost 26,000 years for one full circle. This especially means that the polar star was not exactly in the north e.g. at the time of the Egyptian Pharaohs. And the reason why it does precess is precisely that the earth is not a perfect sphere. First, due to it not being a perfectly rigid body, due to it's rotation it is slightly "fatter" along the equator than on the poles. Moreover, due to inhomogeneities in the mass, and thus in the gravitational field, even without that effect, the earth is not perfectly round, but actually looks like a potato if you artificially increase the differences from the spherical form.
