You can also consider the general case of how teleportation works out for a generic shared state via Choi operators.
Suppose the shared state is some $\rho$, which corresponds to some channel $\Phi$ via $\rho=J(\Phi)$, with $J(\Phi)$ denoting the Choi state of $\Phi$.
More precisely, let $\Phi$ be an arbitrary channel, and let $J(\Phi)\equiv (I\otimes\Phi)\mathbb{P}_\Psi$ be the corresponding Choi operator (again, I'm neglecting normalisation factors). Here I'm using the shorthand notation $\mathbb{P}_\Psi\equiv \mathbb{P}(|\Psi\rangle)\equiv |\Psi\rangle\!\langle\Psi|$ for projectors, and again denoting with $|\Psi\rangle\equiv\sum_i|i,i\rangle$ the maximally entangled state (of relevant dimension).
In such cases one ends up performing some sort of "channel teleportation", meaning that instead of the state $|\psi\rangle$ being faithfully transferred to Bob, he receives the state $\Phi(\mathbb{P}(\bar U_a |\psi\rangle))$ conditionally to Alice measuring the outcome $a$.
We now measure in the basis of maximally entangled states built as $(I\otimes U_a)|\Psi\rangle$, and compute Bob's state conditional to each outcome $a$. In this case we're dealing with generally non-pure states, so we have to change the formalism accordingly, and the expression becomes:
$$
\operatorname{Tr}_{AB}\left\{
\left[\mathbb{P}((I_A\otimes U_a)|\Psi\rangle)\otimes I_C\right](\mathbb{P}_\psi\otimes J(\Phi) )
\right\}
= \sum_{ij} \psi_j \bar\psi_i \operatorname{Tr}_B[ (U_a |i\rangle\!\langle j| U_a^\dagger \otimes I_C) J(\Phi)] \\
= \sum_{ij,k\ell} \psi_j\bar\psi_i
\operatorname{Tr}[U_a |i\rangle\!\langle j| U_a^\dagger |k\rangle\!\langle \ell|] \,\,\Phi(|k\rangle\!\langle\ell|)
= \Phi(\bar U_a \mathbb{P}_\psi \bar U_a^\dagger).
$$
The idea of this expression is to take the input total state $(\mathbb{P}_\psi\otimes J(\Phi) )$, and project the first two spaces onto $(I_A\otimes U_a)|\Psi\rangle$, which in the density matrix formalism is done multiplying with the corresponding projector and partial tracing.
I labeled here with $A,B,C$ the three relevant spaces. So $|\psi\rangle$ leaves in $A$, and $J(\Phi)$ in $BC$ (so in particular note that when writing $\operatorname{Tr}_B$ above, the partial trace is performed on the first of the two spaces in the inner expression).
We get back the standard teleportation case when $\Phi=\operatorname{Id}$ is the identity channel, in which case $J(\Phi)=\mathbb{P}_\Psi$ and we recover the same results as before. Another notable case is when $\Phi$ is a unitary channel that commutes with the unitaries $U_a$, in which case it is again easy to correct the operations on Bob's side, and we perform what is often referred to as "quantum gate teleportation" (cf eg How does quantum gate teleportation differ from state teleportation? and What is quantum gate teleportation?).
The case where the shared state is maximally mixed corresponds to using a fully depolarising channel $\Phi$, and we thus immediately see that no information is sent over.
A closely related post about the relation between Choi and teleportation is https://physics.stackexchange.com/q/270032/58382.
This answer was in part copy pasted from this other answer of mine on physics.SE.