Why does a force parallel to the axis of rotation do not produce torque? One explanation can be that Torque is a cross-product (r x F). I want to know what is its physical interpretation.
 A: Let's take a particular example of a disk rotating around an axis through its centre:

If the component of the torque parallel to the axis is $T_a$ then the disk will start to rotate according to the angular version of Newton's second law:
$$ T_a = I\dot{\omega} \tag{1} $$
where $\dot{\omega}$ is the angular acceleration and $I$ is the moment of inertia. The obvious way to make this rotate is to apply a tangential force:

In this case the magnitude of the torque is just $T=rF$ and the direction of the torque vector is parallel to the axis so $T_a = T = rF$ and equation (1) tells us that the disk will start rotating.
Now suppose we change the direction of the force so it points directly towards the axis:

Obviously this isn't going to make the disk rotate, and equally obviously it's because the force and radial vectors are in a line so $T = \mathbf r \times \mathbf F = 0$ i.e. the torque is zero. That means $T_a$ is zero and equation (1) tells us the disk won't rotate.
The case you ask about is this one:

In this case the torque is non-zero, but the torque vector $\mathbf T = \mathbf r \times \mathbf F$ points out of the screen towards us i.e. the torque vector is at 90º to the axis. That means the component of the torque parallel to the axis, $T_a$, is zero and that's why the disk won't rotate.
