One's naive expectation would be that as the object moves through the medium, it collides with molecules at a rate proportional to $v$. The volume swept out in time $t$ is $A v t$, where $A$ is the cross-sectional area, so the mass with which it collides is $\rho A v t$. The impulse in each collision is proportional to $v$, and therefore the drag force should be proportional to $\rho A v^2$, with a constant of proportionality $C_D$ (the drag coefficient) of order unity.
In reality, this is only true for a certain range of Reynolds numbers, and even in the range of Reynolds numbers for which it's true, the independent-collision picture above is not what really happens. At low Reynolds numbers you get laminar flow and $C_D\propto 1/v$, while at higher Reynolds numbers there's turbulence, and you get $C_D$ roughly constant.
Based on your sketch, the block is always moving along it's long axis, in other words, the velocity is always along the direction of your red vector. This means in the picture on the left, there is only vertical velocity while in the picture on the right, there is both vertical and horizontal velocity. This is what you have described, I am only summarizing to make sure the rest of my answer makes sense and is consistent with this understanding.
Now, you claim the drag coefficient of the case on the left is known yet the drag coefficient for the case on the right is unknown. To clarify a point, the drag coefficient is a scalar, not a vector as you claimed it to be. And in this case, the drag coefficient is actually the same in both cases if the block is moving along the vectors shown. Why is this the case?
Consider the area vector of each face to be the area of the face times the surface normal, ie.:
$$\vec{A} = A\cdot\hat{n}$$
In your example on the left, the area normal is in the $\hat{j}$ or Y-normal direction and so is the velocity. This gives you the drag coefficient you cite.
In your example on the right, you now have a component of the area vector in the $\hat{i}$ direction and a component in the $\hat{j}$ direction. But your velocity also has components in both those directions.
You are correct to say that the area increases in the $\hat{j}$ direction -- you now have both a portion of the short side and a portion of the long side exposed to the Y-direction. However, the block is still moving normal to the short side along the red vector, which means the area vector for the long sides are perpendicular to the red vector and that area does not contribute to the so called "frontal area" of the block.
Best Answer
Yes there will be a drag torque opposite the direction of spin. The name for this seems to be viscous torque. According to this paper, the viscous torque on a spinning sphere of radius $R$ in a fluid with viscosity $\eta$ spinning with constant angular velocity $\vec\Omega$ is $$ \vec\tau = -8\pi R^3\eta\vec\Omega $$ The paper goes on to describe how to generalize this relationship to the case of a sphere rotating with arbitrary time-varying angular velocities (under some other assumptions).
If we, however, assume that to good approximation, the viscous torque is linear in the instantaneous angular velocity, then the sphere under the influence of a constant external torque $\vec\tau_0$ will, as you point out, reach terminal angular velocity when the external torque and viscous torque are equal in magnitude but opposite in direction $$ \vec\tau=-\vec\tau_0 $$ This follows from the fact that the net torque $\vec\tau_{\mathrm{net}}$ on the sphere (undergoing fixed axis rotation) is related to its angular acceleration $\vec\alpha$ and moment of inertia $I$ about it's rotation axis by $$ \vec\tau_\mathrm{net} = I\vec\alpha $$ so if the net external torque is zero (which happens when the viscous torque equals the other external torque), then $$ \vec\alpha = 0 $$ the angular acceleration vanishes, so the sphere will no longer speed up in its spinning.