Admittedly, I never found an official reference that confirms some of the things I mention below, which therefore are just my educated guess of the topic.
In particular, if someone has further mathematical insight about why it is legitimate (or illegitimate!) to write the Hilbert spaces as I'm about to do, it is welcome to correct me;
The position eigenbasis $\{|x,y,z\rangle\}$, eigenvectors of Hermitian position operators $\hat X$, $\hat Y$, $\hat Z$, is a complete basis for the Hilbert space of a 3-dimensional system, $\mathcal H_{\vec x}$, which can be considered as the tensor product of the three Hilbert spaces $\mathcal H_{x}\otimes\mathcal H_{y}\otimes\mathcal H_{z}$, each of which is a $L^2(-\infty,+\infty)$ and is spanned by one of the $\{|x\rangle\}$, $\{|y\rangle\}$, $\{|z\rangle\}$.
From a general perspective, the three Cartesian coordinates are not the unique possible choice, and we could think to translate a position eigenket in spherical coordinates as
$$|x,y,z\rangle\rightarrow |r,\theta,\phi\rangle$$
In this case, similarly, we postulate the existence of other three Hermitian operators $\hat R$, $\hat \Theta$ and $\hat \Phi$:
commuting with one another,
with spectra respectively $[0,\infty)$, $[0,\pi]$, $[0,2\pi]$,
with eigenbasis $\{|r\rangle\}$, $\{|\theta\rangle\}$, $\{|\phi\rangle\}$, spanning respectively a $L^2[0,\infty)$, a $L^2[0,\pi]$ and a $L^2[0,2\pi]$; the total Hilbert space now can be written as a $\mathcal H_{r}\otimes\mathcal H_{\theta}\otimes\mathcal H_{\phi}$.
$|r,\theta,\phi\rangle=|r\rangle\otimes |\theta\rangle\otimes|\phi\rangle$ is their simultaneous eigenvector with eigenvalues $r$, $\theta$, $\phi$.
With this foreword, let's now focus on the action of the angular momentum operator $\hat L=\hat X\times\hat P$:
by definition, it is the generator of infinitesimal rotations about some axis, meaning that if it acts on $|r,\theta,\phi\rangle$, it can at most modify $\theta$ and $\phi$ but will leave $r$ invariant.
As this happens for any $|r,\theta,\phi\rangle$, this means that the action of $\hat L$ on the Hilbert space $\mathcal H_{r}\otimes\mathcal H_{\theta}\otimes\mathcal H_{\phi}$ is the identity on the subspace $\mathcal H_{r}$, and some nontrivial operator on the other subspace.
In practice, we can say that $\hat L$ only acts on $\mathcal H_{\theta}\otimes\mathcal H_{\phi}$, so that's where the nontrivial part of its eigenvectors will live.
As $\hat L$ is Hermitian, its eigenvectors form a basis for the Hilbert space where it operates, so together with $\{|\theta,\phi\rangle\}\equiv\{|\theta\rangle\otimes|\phi\rangle\}$ we have another choice, which is $\{|l,m\rangle\}$.
The so called spherical harmonics are defined as
$$Y_l^m(\theta,\phi)\equiv \langle \theta,\phi|l,m\rangle$$
and can be seen either as the matrix of the change of basis between $|\theta,\phi\rangle$ and $|l,m\rangle$, or as the eigenfunctions of $\hat L$, i.e. the projection on the coordinate space of the eigenstates $|l,m\rangle$ of $\hat L$.
In some sense, as $\hat L$ generates "translations" in the angle space, we can say that the relationship between $|\theta,\phi\rangle$ and $|l,m\rangle$ is very similar to the one between $|x,y,z\rangle$ and $|p_x,p_y,p_z\rangle$, and the spherical harmonics are then the equivalent of the plane waves $\langle x|p\rangle\propto e^{ipx/\hbar}$.
They also share the analogy to be a complete set of functions in their space of definition.
Finally:
about the $\hat n$, specifying $\theta,\phi$ is equivalent to specify a direction $\hat n$, so you can use $|\hat n\rangle$ or $|\theta,\phi\rangle$ interchangeably.
about $|n,l,m\rangle$: as someone has already pointed out, that $n$ is not the direction $\hat n$, but the quantum number related to the energy.
If the Hamiltonian commutes with angular momentum operators $\hat{\vec L}^2$ and $\hat L_z$, then you can build a simultaneous eigenbasis $|E,l,m\rangle$ which is usually written as $|n,l,m\rangle$ if the energy is quantized and labeled by an index $n$, $E_n$.
But the operator $\hat H$, differently from angular momentum, acts on the full Hilbert space, so $|l,m\rangle$ or $|\theta,\phi\rangle$ are not enough to fully expand it and we need to resort to the full $|r\,theta,\phi\rangle$. The usual wavefunction is written as:
$$\langle r,\theta,\phi|n,l,m\rangle=\psi_{nlm}(r,\theta,\phi)$$
which thanks to the commutativity of the Hamiltonian with the angular momentum further separates as
$$\langle r,\theta,\phi|n,l,m\rangle=R_{nl}(r)Y_l^m(\theta,\phi)$$
