To answer your question, the angular velocity of any constituent particle in a rotating rigid body is calculated from a perpendicular displacement from the rotation axis, and not from an arbitrary choice of your coordinate system's origin.
If it were the latter, then any angular velocity could be ascribed to that particle by a suitable choice of origin. And since the dynamics of a rotating rigid body are independent of coordinate system, the angular velocity vector cannot be so constructed.
More intuitively, angular mechanics is at its essence an evaluation of motion in terms of pure circular trajectories, as opposed to linear mechanics in which motion is evaluated in terms of straight-line trajectories. As such, the angular velocity vector encodes the idea that the particle moves in a circle in its very construction. Were you to take a single particle at position $\vec r$ and imparted a velocity $\vec v$, then without constraint the particle would move linearly (along a trajectory $\vec{s}(t) = \vec r + t\,\vec v$). If the particle were fixed to stay at locus $r$ from the origin $O$, then it would rotate about $O$. The rate of rotation is given by the angular velocity $\vec\omega = \vec r \times \vec v/ r^2$, which simply calculates the component of the velocity vector perpendicular to $\vec r$ as the velocity that is pertinent to circular motion: $\vec \omega = \vec r \times \vec v/ r^2 = \frac{v \sin\theta}{r} \hat\Sigma$ (where $\hat \Sigma$ is a basis vector perpendicular to the plane of motion; in fact, you can interpret this $\hat\Sigma$ to define your plane of motion). Thus, whenever angular velocities are calculated it is always the displacement from that centre of the circle that the particle will move along that is used (or, equivalently, the perpendicular displacement from the rotation axis), since the whole point of $\vec \omega$ is to describe motion about that point.