Is angular velocity invariant under SR? Is angular velocity invariant under special relativity, i.e. do all observers in any relative inertial frames measure same value for angular velocity of a system? If not, what is its expression?
 A: For a classical particle angular momentum is often defined as $\vec{r}\times\vec{p}$ (but it is actually a tensor even in non relativistic classical physics.)
In special relativity we generalize it as  a tensor. When expressed in Contravariant components it is given by
$$M^{\alpha \beta }=X^{\alpha }P^{\beta }-X^{\beta }P^{\alpha }$$
Once we define like this it is conserved in Special Relativity.

do all observers in any relative inertial frames measure same value for angular velocity of a system?

No. In special relativity we can transform from one observer to another by multiplying Lorentz matrices like this
\begin{aligned}{M'}^{\alpha \beta }&={X'}^{\alpha }{P'}^{\beta }-{X'}^{\beta }{P'}^{\alpha }\\&={\Lambda ^{\alpha }}_{\gamma }X^{\gamma }{\Lambda ^{\beta }}_{\delta }P^{\delta }-{\Lambda ^{\beta }}_{\delta }X^{\delta }{\Lambda ^{\alpha }}_{\gamma }P^{\gamma }\\&={\Lambda ^{\alpha }}_{\gamma }{\Lambda ^{\beta }}_{\delta }\left(X^{\gamma }P^{\delta }-X^{\delta }P^{\gamma }\right)\\&={\Lambda ^{\alpha }}_{\gamma }{\Lambda ^{\beta }}_{\delta }M^{\gamma \delta }\\\end{aligned}
Read more about it here in Wikipedia or in chapter 14 of this book by Michael Tsamparlis
