The Hamiltonian
Just as for vibrations we have the harmonic oscillator approximation, for rotations we often use the rigid rotor approximation, where bond lengths are fixed. Recall the rigid-rotor Hamiltonian (in this case the kinetic energy operator) for a diatomic, which is often written as follows:
$$\tag{1}
\hat{H} = \hat{T} = \frac{J_x^2}{2I_x} + \frac{J_y^2}{2I_y} + \frac{J_z^2}{2I_z},
$$
where $\mathbf{\hat{J}} = \mathbf{\hat{r}}\times \mathbf{\hat{p}}$ is the angular momentum operator, and $(I_x,I_y,I_z)$ are the diagonal elements of the moment of inertia tensor when its written in the $(\mathbf{x},\mathbf{y},\mathbf{z})$ coordinate system..
The Hamiltonian for polyatomics can be written in the same way, but instead of using the lab-fixed frame which has the fixed axes $(\mathbf{x},\mathbf{y},\mathbf{z})$, it's easier to use a body-fixed or molecule-fixed frame with axes that rotate along with the molecule: these axes (known as the "principal axes") will be the eigenvectors $(\mathbf{a},\mathbf{b},\mathbf{c})$ of the moment of inertia tensor, and the corresopnding eigenvalues are denoted by $(I_a,I_b,I_c)$. The Hamiltonian (kinetic energy operator) is now:
$$\tag{2}
\hat{H} = \hat{T} = \frac{J_a^2}{2I_a} +\frac{J_b^2}{2I_b}+\frac{J_c^2}{2I_c}.
$$
The solution
As you noted correctly in your comment we have three quantum numbers, let's call them $(j,k,m)$.
To solve the Schroedinger equation for this Hamiltonian, we can model the wavefunction in terms of the three Euler angles:
$$\tag{3}\psi_{jkm}(\theta,\chi,\varphi) \equiv |jkm\rangle = P_{jkm}(\theta) e^{\textrm{i}k\chi}e^{\textrm{i}m\varphi},
$$
which has the following eigenvalues for various operators:
\begin{align}
\hat{J}^2|jkm\rangle &= j(j+1)|jkm\rangle, ~~~~~\left(\hat{J}^2 \equiv \hat{J}_x^2 + \hat{J}_y^2 + \hat{J}_z^2 = \hat{J}_a^2 + \hat{J}_b^2 + \hat{J}_c^2 \right), \tag{4}\\
\hat{J}_z|jkm\rangle &= m|jkm\rangle,\tag{5}\\
\hat{J}_c|jkm\rangle &=k|jkm\rangle,\tag{6}\\
\end{align}
and the following results when raising your positions on the $m$ and $k$ ladders respectively:
\begin{align}
\hat{J}_{m\pm}|jkm\rangle &= \sqrt{(j \mp m)(j \pm m +1)}|j,k,m\pm1\rangle, ~~~~ \hat{J}_{m\pm} \equiv \hat{J}_x \pm \textrm{i}\hat{J}_y,\tag{7}\\
\hat{J}_{k\pm}|jkm\rangle &=\sqrt{(j \mp k)(j \pm k +1)}|j,k\pm1,m\rangle, ~~~~~~~ \hat{J}_{k\pm} \equiv \hat{J}_a \mp \textrm{i}\hat{J}_b.\tag{8}\\
\end{align}
Now with the following definitions:
\begin{align}
\alpha &\equiv \frac{1}{4}\left( \frac{1}{I_a} + \frac{1}{I_b}\right), \tag{9}\\
\gamma &\equiv \frac{1}{8}\left( \frac{1}{I_a} - \frac{1}{I_b}\right), \tag{10}\\
\beta &\equiv \frac{1}{2}\left( \frac{1}{I_c} - \frac{1}{2}\left(\frac{1}{I_a} + \frac{1}{I_b}\right)\right), \tag{11}
\end{align}
we can write Eq (2). in the following algebraicly equivalent form for which our wavefunction ansatz $|jkm\rangle$ is already an eigenvector for many cases:
$$
\hat{H} = \alpha \hat{J}^2 + \beta \hat{J}_c^2 + \gamma ( {\hat{J}^+}^2 +
{\hat{J}^-}^2 ).\tag{12}
$$
Now we're ready to get the solutions for various cases based on the symmetry present in the molecule:
Spherical top molecules (e.g. $\ce{CH4}$):
All eigenvalues of the moment of inertia tensor will be equal to each other: $I_a = I_b = I_c \equiv I$, so we have $\beta = \gamma = 0$ and $\alpha=\frac{1}{2I}$:
\begin{align}
H &= \frac{ J^2}{2I} \tag{13}\\
H|jkm\rangle &= \frac{j(j+1)}{2I}|jkm\rangle \tag{14}\\\\
E&= \frac{ j(j+1)}{2I} \tag{15}\\
|\psi\rangle&=|jkm\rangle.\tag{16}\\
\end{align}
Symmetric top molecules (e.g. $\ce{NH3}$):
Two eigenvalues of the moment of inertia tensor will be equal to each other, so if we pick them to be the ones with eigenvectors $\mathbf{a}$ and $\mathbf{b}$ we will get $\gamma=0$:
\begin{align}
H &= \alpha J^2 + \beta J^2_c \tag{17}\\
H|jkm\rangle &= \alpha j(j+1)|jkm\rangle + \beta k^2|jkm\rangle \tag{18}\\
&= \left( \alpha j(j+1) + \beta k^2 \right) |jkm\rangle \tag{19}\\\\
E&= \alpha j(j+1) + \beta k^2 \tag{20}\\
|\psi\rangle&=|jkm\rangle.\tag{21}\\
\end{align}
Linear molecules (e.g. $\ce{CO2}$):
The situation is the same as for spherical tops.
Asymmetric top molecules (e.g. $\ce{H2O}$):
Diagonalizing the matrix is not as simple, but it's not too bad if you use $J_\pm$ to build the Hamiltonian for each $(j,m)$ pair and for $k$ from $-j$ to $j$.
General non-symmetric molecules (e.g. $\ce{CHBrClF}$):
The eigenvalues and wavefunctions would typically be obtained by numerically diagonalizing submatrices of size $(2j+1)$ × $(2j+1)$ .
Some helpful resources:
Much of what I wrote here can be found here, and I also very much liked this and for the relation to the diatomic case: this and this.
