I'm going to disagree with the other answers: I think that the angular momentum of macroscopic "classical" objects is not quantized.
Consider an automobile tire spinning in a wheel well. On the tire is a device that triggers whenever a certain point on the wheel crosses a certain point on the well, adding one to an internal counter if it passes it clockwise and subtracting one if it crosses it counterclockwise. (Alternately, you could dispense with the wheel well and say that the device tracks its own position by inertial navigation.) The state of this system can be described by a value $θ\in\mathbb R$, where the current value of the counter is $\lfloor θ/2π\rfloor$ and the angle of the wheel is $θ\text{ (mod }2π\text{)}$. In the absence of external forces, the Hamiltonian of the system is essentially that of a free particle in $\mathbb R$, and the spectrum of angular momenta is continuous just like the free particle's momentum spectrum.
That's a 2+1 dimensional system. In 3+1 dimensions, there's the Dirac belt trick to worry about. Does it matter? I don't think so. There's no reason to limit the device to holding a single integer, or to being reversible. It could simply store the entire history of its orientation readings internally, or broadcast them by radio, indelibly recording them in the universal wave function. That's a very noncompact state space, and it's an accurate enough model of bodies like the earth.
The angular momentum operator on this monstrosity obviously violates the assumptions of any proof of the quantization of angular momentum, but that's no reason not to call it angular momentum. We do call it angular momentum, and it's what the question was about.
In response to comments I'll try to clarify my answer.
These are quantum systems, but the earth system is "classical" in the sense of being a quantum system with emergent classical behavior.
The reason that high temperature systems behave classically is that they constantly leak which-path information into the environment. If you do a double-slit experiment with the earth, it will emit different patterns of light going through one slit than through the other. You can literally see which slit it goes through, but even if you don't look, the which-path information is there in the patterns of light, or in patterns of heat if the light is absorbed by the walls of the lab, and that's all that's necessary to make the final states orthogonal and destroy the interference pattern.
It's sometimes said that you can't see an interference pattern in the earth double-slit experiment simply because its de Broglie wavelength is so small. That would be correct for a supermassive stable particle that doesn't radiate, but it's wrong for the earth. For the earth there's no interference pattern at all, for the same reason there's no interference pattern when there's a detector at one of the slits. Earth's thermal radiation is the "detector".
The case of rotation is similar. Different rotations are different paths through the state space (it's the state space, not physical 3D space, that the wave function is defined on and which matters here). If you consider two different paths ending in the same physical orientation (analogous to the same position on the screen in the double-slit case), these paths will interfere if no information about which path was taken is recorded anywhere. In the case of the earth, this means they'll interfere if there's no way for anyone to tell whether the earth rotated around its axis or not. If there's any record of it – if any animals remember the day-night cycle, or don't but could in principle, or if aliens see it rotate through a telescope, or don't but could in principle – then there's no interference.
The proof that angular momentum is quantized depends on the compactness of the space of orientations. This is fine if the space of orientations is the phase space, i.e., if the system is memoryless. If it has a memory, rotating the system through $2π$ or $4π$ doesn't leave it in the same state as not rotating it.
The tire example in the second paragraph may have been a mistake since it seems to have only caused confusion. But it's a perfectly good quantum system in the abstract, and its state space is $\mathbb R$, not $S^1$.