The answer turned out to be much more complex than I thought. Here is the dump of my day worth of research on this subject. Below is, of course, simplified view and doesn't consider many other effects in fluid dynamics such as turbulent flows, non-Newtonian fluids etc.
Two types of angular drag
The damping of angular motion in fluid happens due to two main types of drag effects: (1) tangential velocity and (2) normal velocity to the surface. The former is known as shear stress however there doesn't seem to be agreed upon name for the later. To keep in line with standard textbook "Introduction to Fluid Mechanics" p.160, sec 9.3.5, we will refer to it as friction torque.
To get an intuitive idea, imagine a cylinder placed at origin with its height aligned with z-axis and circular face on x-y plane. If you rotate this cylinder around x or y axis with angular velocity $\omega$ then fluid hitting the shaft would be normal to its surface. If you were to rotate this cylinder around z axis with angular velocity w then there shaft surface is experiencing only tangential velocity.
Linear drag
First lets revisit linear drag because we will use it to compute angular friction drag later. Linear drag force is given by,
$$F_{lin} = \frac{1}{2} \rho v^2 . C_{lin} . A$$
A fun fact is that majority of textbooks simply states this equation as "fact" never bothering to examine it closely. I wanted to understand how this equation came to existence and so far the intuitive explanation is that force exerted on body moving with relative velocity $v$ can be given in terms on pressure on its exposed area:
$$F_{lin} \propto P A$$
Now the pressure exerted can be given as kinetic energy of the fluid,
$$P = \frac{1}{2} \rho v^2$$
Above is known as dynamic pressure. So if you were wondering how did that $\frac{1}{2}$ slipped in to equation, it traces back to kinetic energy definition. You can also find entertaining introduction to this equation in essay "The velocity dependence of Aerodynamic Drag" by Long and Weiss.
Friction Torque
Its fairly easy to compute friction torque. Each tiny surface area $ds$ of body located at vector $r$ experiences the linear velocity $v = r \times \omega$ where $w$ is angular velocity. So you just take that linear velocity, plug in to equation for linear drag to get the force $dF$ exerted on that tiny area $ds$. Then you get torque $dT = dF \times r$. Finally integrate over entire exposed surface and you get total friction torque.
Here's the worked example for cylinder (also see: Introduction to Fluid Mechanics, sec 9.3.5, p. 160)
First compute the frictional torque for the circular top and bottom discs with radius $r$. Note that $r \times \omega$ is simply $rw$ because fluid is hitting right on surface normal.
$$dF_{disc} = dP_{disc} . C_{lin} . dA_{disc} dr$$
$$dF_{disc} = \frac{1}{2} \rho (rw)^2 . C_{lin} . 2\pi r . dr$$
$$T_{disc} = \int_{r=0}^{r=r_0} r . dF_{disc} = \frac{1}{5} \pi C_{lin} \rho \omega^2 r_0^5$$
Now lets find the frictional torque experienced by shaft part of the cylinder with height $h$. This time around there is no need to do integration because $r$ is fixed over the surface of the shaft.
$$F_{shaft} = P . A . C_{lin}$$
$$F_{shaft} = \frac{1}{2} \rho (r_0 \omega)^2 . 2\pi r_0 h . C_{lin}$$
$$T_{shaft} = \pi . \omega^2 r_0^4 h \rho C_{lin}$$
Depending on axis of angular velocity you may need to use shaft and/or disc torques and add them to get total frictional torque. Also note that $F_{shaft}$ might be only half because only half the surface area is experiencing fluid in surface normal.
Shear Stress
Now lets move on to this other variety. This one is particularly difficult to compute because velocity from body is transferred to surrounding fluid via tangential component at a point on body. So there is this "boundary layer" formed surrounding body which is rotating at same speed as body at the touch point but then its velocity reduces linearly as go you go further out from the body. The big issue is that equations assume limited width of boundary layer and that width appears in denominator of force. That means if you don't get width right, force value can technically become very error prone and even arbitrary! I would suggest to check Wikipedia or here to view the related equations. In the book "Viscous Fluid Flow", 2nd ed, by White, p112 there is a case for cylinder that has boundary width of infinite size (which also requires cylinder to be of infinite height) that would yield the drag force as,
$$F_{shear} = 2 \pi \mu \omega r h$$
where $\mu$ is dynamic viscosity coefficient which for air at room temperature is $1.8 \times 10^{-5}$.
(see derivation here)
End notes
I'm looking to use all these in practice but the calculations for arbitrary real-world shapes might certainly get hard. Also notice that we don't take in to account all the effects that different shapes likes concave or wings etc can produce. From looking at purely applied point of view, it seems that drag due to shear would be very small because of not only very small constant $\mu$ for air but also the fact it depends only on first power of $\omega$ and second power of $r$. So I might ignore it all together for practical purposes as I don't have any shapes that will amplify its effect. For frictional drag, one approximation might be to compute area "component" in each axis to form "area vector" and then do coefficient wise multiplication with velocity vector (in body frame) twice. Here velocity vector may simply be approximated by multiplying angular velocity with average encompassing radius of body in axis's direction.
Another quick note: So how does physics simulation engines handle angular drag computation? So far I have see that they take easy way out: Just define some constant like 0.01 and reduce angular velocity by that amount each time step. This eliminates surprises like forever rotating body but this just simply isn't right, obviously.
Word of caution: I'm not expert on fluid dynamics so please do not view above as an official statement from an expert and please feel free to comment if you see any mistakes! It would be super awesome if some expert in the field can validate this answer, of course!
