First: I'm not familiar with the book you're talking about. And it's not clear exactly what you mean in a mathematical sense by a tensor being "applied" to the torque "before applying" it to angular velocity. If you mean something like T/I = delta-w, that's true.
am I missing something?
I think so. In general, you may get a better feeling for torque, angular momentum and angular velocity by considering those as parallels with classical Newtonian linear force, linear momentum and linear velocity. I.e., stuff like linear momentum P = m * v, and F = m * a. So, angular momentum L = I * w, and Torque = I * dw/dt (angular acceleration).
rotational momentum is conserved
That applies to a system, not a single object. Obviously, an object's angular momentum can change through the application of torque. But, for the torque applied, an equal and opposite torque must be applied to something else in the system. That's similar to linear Newtonian physics: linear momentum is conserved for a system. The linear momentum of a rigid object, if no external force is applied, remains constant. If an external force is applied, an equal and opposite force must be applied to something else. In that case, both the rigid object and the "something else" must be considered as the system in question, and the momentum of the system remains constant.
... if no torque is applied, the body will spin around the same axis, at the same speed. If we use the classic example of an ice skater, this means that drawing in your arms won't make you spin faster.
Your assumptions are incorrect. If no external torque is applied, then there will be no angular acceleration due to external torque. Drawing your arms in changes the moment of inertia. So, because L = I*w, and I has changed, w must change. That is, I is the integral of mass * radius-from-axis. Mass is conserved, so when the radius decreases, the moment of inertia decreases. If I decreases, w must increase. The change in w results from the torque that results from the force the skater applies to the arms to draw them in. Force-at-a-distance is torque.
I suspect you know all that. Perhaps if you provide the exact context of your question, a better explanation can be provided.