Yes, angular momentum is conserved if you do the problem correctly.
If you assume the ball bounces back along exactly the same path it followed before the collision, there are three degrees of freedom: the velocity of the ball, the velocity of the rod, and the rotational rate of the rod. There are three constraints: conservation of linear momentum in the direction of motion of the ball, conservation of angular momentum, and conservation of kinetic energy. You seem to understand this.
From there, it's unclear what your approach was. How did you get the particular numbers you cite? There are infinitely-many ways to conserve linear momentum and kinetic energy. Conserving just those two applies two constraints to a problem with three degrees of freedom, so there is an entire one-dimensional manifold of solutions. IE you could give the ball any velocity you like (up to a maximum), then choose the rod's velocity and rotation rate to fit the two constraints. If you simply pick one of these solutions at random, then it is very unlikely to conserve angular momentum. You must use all three constraints to solve the problem.
You should do this an confirm that the correct figures are
$$v_{ball} = -\frac{66}{7} m/s$$
$$v_{rod} = \frac{22}{7} m/s$$
$$\omega = \frac{132}{7}s^{-1}$$
First, angular momentum isn't measured about an axis. It's measured about a point.
Second, well, of course the angular momenta about different points will be different in general. But they will each be conserved -- there's no need for the point to be in the axis of rotation or even in the same galaxy as the rotating object you're interested in.
Now, about your example. The total angular momentum, on the three block system, is definitely conserved. Why wouldn't it be? What you're perhaps missing is the fact that there will be in general an angular momentum associated with the 3rd block as well, even if it's traveling on a straight line. For example, say you have a block with momentum $\vec{p}$ traveling in a straight line at constant velocity until it hits the block at one end of the string (at the point of closest approach to the center of mass). Say the string has length $2\ell$. Now let's calculate the angular momentum of this third block about the center of mass of the two rotating blocks when it's at a distance $r$ from that center of mass.
$$\vec{L} = \vec{r} \times \vec{p} = rp \sin \theta \hat{z} = rp \frac{\ell}{r}\hat{z} = rp \hat{z} $$
(The $\hat{z}$ is just my choice in how to orient the system in space).
Notice now that $L$ does not depend on the distance $\vec{r}$. It's a constant of the motion, as promised, so long as $\vec{p}$ itself is conserved. Notice that there was nothing special in this derivation about the center of mass: it could've been literally any point and the conclusion still would be valid.
Now you can work out what will happen in the collision, assuming that linear momentum is conserved, and you'll prove explicitly that angular momentum is conserved in this process.
Best Answer
First off, there is no reason to expect that any of the conservation laws apply to the rod. A moving particle collides with the rod, and the rod has constraints that act on it to keep one end fixed. The collision and those constraint forces are external forces, some of which result in external torques. The conservation laws don't apply to the rod. They apply to the rod+particle+Earth system.
In general,
Secondly, you are ignoring that even point masses can have non-zero angular momentum. Angular momentum is always measured with respect to a point, not an axis. The angular momentum of a point mass is easily computed: It is $\vec L = m \vec r \times \vec v$, where $m$ is the mass of the point mass, $\vec r$ is the displacement vector from the central point to the point mass, and $\vec v$ is the velocity of the point mass. When viewed from the right perspective, the rod+particle system does conserve angular momentum. This "right perspective" is one in which the constraint forces on the rod exert zero torque.