Let's see, how mathematically the angular velocity comes about. To this end think of a rigid rod with one end fixed and free to rotate around this point. Make ##\vec{n}(t)## the vector from the fixed end to the other end. Then it's clear that
$$\vec{n}(t) = \hat{D}(t) \vec{n}_0,$$
where ##\vec{n}_0=\vec{n}(0)## is the vector at time ##T=0##, and ##\hat{D}(t)## is a rotation. In the following I consider ##\vec{n}## as the column vector of its components wrt. a Cartesian basis. Then ##\hat{D}(t)## is a orthogonal matrix, fulfilling ##\hat{D}^{\mathrm{T}} \hat{D}=\hat{D} \hat{D}^{\mathrm{T}}=\hat{1}## with ##\mathrm{det} \hat{D}=1##.
Now let's take the time derivative:
$$\dot{\vec{n}}(t)=\dot{\hat{D}}(t) \vec{n}_0 = \dot{\hat{D}}(t) \hat{D}^{\text{T}}(t) \vec{n}(t). \qquad (*)$$
Now from the orthogonality of ##\hat{D}## we have
$$\dot{\hat{D}} \hat{D}^{\text{T}} = -\hat{D} \dot{\hat{D}}^T = - \left ( \dot{\hat{D}} \hat{D}^{\text{T}} \right)^{\text{T}},$$
i.e., ##\hat{\Omega}=\dot{\hat{D}}(t) \hat{D}^{\text{T}}(t)## is an antisymmetric matrix, and we thus can write
$$\Omega_{jk}=\epsilon_{jlk} \omega_l,$$
and thus from (*)
$$\dot{n}_j=\Omega_{jk} n_k = \epsilon_{jlk} \omega_l n_k = (\vec{\omega} \times \vec{n})_j,$$
or
$$\dot{\vec{n}}=\vec{\omega} \times \vec{n}.$$
One calls ##\vec{\omega}## the angular velocity. The change of the vector ##n## during a small time ##\mathrm{d} t## is
$$\mathrm{d} \vec{n} = \mathrm{d} t \vec{\omega} \times \vec{n}.$$
From the geometrical meaning of the dot product that means that the infinitesimal rotation is around an axis given by the direction ##\vec{\omega}/|\vec{\omega}|##, implying the direction of rotation to be according to the right-hand rule, and the infinitesimal rotation angle is ##|\vec{\omega}| \mathrm{d} t##.
Of course you can decompose ##\vec{\omega}## in terms of an arbitrary (Cartesian) basis, and the infinitesimal rotation can be seen as composed of several infinitesimal rotations around the corresponding axes, but still it's just a rotation around one given axis.
