For a point, you do talk about its velocity. By abuse of language, for an axis aligned with $\vec{n}$, I've seen people refer to angular velocity of the point about the axis as $\dot{\theta}\hat{n}$, that is, if you express the coordinates of the point in a cylindrical coordinate system with coordinates $(\rho, \theta, z)$ in which $\hat{n}$ is aligned with the cylindrical axis.

As far as I understand, this is not standard and generally rather ambiguous.

There is a physical way to imagine this, though. Imagine a/an (infinitely extended) rigid body which can perform only screw motion and/or rotation about the given axis, in a manner such that the given moving point is stationary wrt to this rigid body. Then, the "angular velocity of the point defined wrt the axis" is identical to the angular velocity of the aforementioned rigid body.

However, I'd say it's best to avoid talking about angular velocity of a point defined this way, since it depends on the choice of the origin (even in the same reference frame).

On the other hand, for a rigid body, there is a very precise notion of angular velocity, which doesn't depend on the choice of the origin as long as you stick to the same frame.

To define the angular velocity of a rigid body, you need to know the velocity field $\mathbf{v}(\mathbf{r})$ in a given frame. The rigidity constraint then implies that in a Cartesian coordinate system the velocity field can be decomposed as,

$$\mathbf{v}(\mathbf{r}) = \mathbf{v}_0+\mathbf{\omega}\times(\mathbf{r}-\mathbf{r}_0).$$

It can be shown that $\omega$ is independent of the origin of the coordinate system (eg. cf. Landau-Lifshitz Mechanics).

So, you can see that $\mathbf{\omega}$ is a quantity that comes out of the collective motion of all points on the rigid body. And this is the standard, commonly-accepted definition of angular velocity for a rigid body.

For a point, you do talk about its velocity. By abuse of language, for an axis aligned with $\vec{n}$, I've seen people refer to angular velocity of the point about the axis as $\dot{\theta}\hat{n}$, that is, if you express the coordinates of the point in a cylindrical coordinate system with coordinates $(\rho, \theta, z)$ in which $\hat{n}$ is aligned with the cylindrical axis.

As far as I understand, this is not standard and generally rather ambiguous.

There is a physical way to imagine this, though. Imagine a/an (infinitely extended) rigid body which can perform only screw motion and/or rotation about the given axis, in a manner such that the given moving point is stationary wrt to this rigid body. Then, the "angular velocity of the point defined wrt the axis" is identical to the angular velocity of the aforementioned rigid body.

However, I'd say it's best to avoid talking about angular velocity of a point defined this way, since it depends on the choice of the origin (even in the same reference frame).

On the other hand, for a rigid body, there is a very precise notion of angular velocity, which doesn't depend on the choice of the origin as long as you stick to the same frame.

To define the angular velocity of a rigid body, you need to know the velocity field $\mathbf{v}(\mathbf{r})$ in a given frame. The rigidity constraint then implies that in a Cartesian coordinate system the velocity field can be decomposed as,

$$\mathbf{v}(\mathbf{r}) = \mathbf{v}_0+\mathbf{\omega}\times(\mathbf{r}-\mathbf{r}_0).$$

It can be shown that $\omega$ is independent of the origin of the coordinate system (eg. cf. Landau-Lifshitz Mechanics).

So, you can see that $\mathbf{\omega}$ is a quantity that comes out of the collective motion of all points on the rigid body. And this is the standard, commonly-accepted definition of angular velocity for a rigid body.

ABUSE OF LANGUAGE

The question talks about "the angular velocity of a rigid body about the 'axis'" - this is a meaningless phrase if you do not specify the frame of reference. For instance, one could switch to another frame rotating about the axis with some angular velocity $\Omega$. If you do so, the angular velocity of the rigid body about the axis would change, even though the said 'axis' wouldn't acquire any additional velocity in the changed frame of reference.

Long story short: An axis doesn't specify a reference frame. You need two more perpendicular axes (and what they are doing) to complete the story.

I believe the question, as it stands, has plenty of abuse of language and is fairly ambiguous.