Let's assume we have a register of qubits present in a mixed state $$\rho = \sum_i^n p_i|\psi_i\rangle \langle \psi_i|$$
and we want to teleport $\rho$ through a random pure state $|\phi\rangle$. What would be the result of this teleportation?
Additionally, what would happen if $|\phi\rangle$ is one of the pure states $|\psi_i\rangle$, $\rho$ is composed of?
I do struggle a little with the density matrix formalism of QC, therefore, I could use some help here.
The teleportation should behave just the same with a mixed state as it does with a pure state. I'm going to assume a bit of familiarity with how teleportation works for pure states, as you can find many resources addressing that problem.
We can describe the task of teleportation as follows: systems $M$, $A$, and $B$, our goal is to prepare the state $\rho_M$ (subscript denotes that its currently stored in $M$) in system $B$, using only local operations over $MA$ plus some preexisting entanglement between $AB$. Its simplest to begin with a Bell state prepared over $AB$: $$\tag{1} |\Phi_0\rangle_{AB} = \frac{1}{\sqrt{2}}(|00\rangle_{AB} + |11\rangle_{AB}) $$
and so the initial state defined over $MAB$ is given by: $$\tag{2} \rho_M \otimes |\Phi_0\rangle\langle \Phi_0|_{AB}= \left(\sum_{i=1}^n p_n |\psi_i\rangle\langle \psi_i|_M \right)\otimes |\Phi_0\rangle\langle \Phi_0|_{AB} $$
When one works out teleportation for a pure state $|\psi_i\rangle$ its straightforward (but tedious) to show that $$\tag{3} |\psi_i\rangle_M\otimes |\Phi_0\rangle_{AB} = \frac{1}{2}\sum_{\ell=0}^3 |\Phi_\ell\rangle_{MA} \otimes \sigma_\ell|\psi_i\rangle_B $$
where $\{\sigma_\ell\}$ are the pauli operators $\{I,X,Y,Z\}$ and $\{|\Phi_\ell\rangle \}$ are Bell basis states$^1$ for $\ell=0,1,2,3$. But since the tensor product is linear we can just substitue (3) into (2) to rewrite the initial state before teleportation as: \begin{align}\tag{4} \rho_M \otimes |\Phi_0\rangle\langle \Phi_0|_{AB} &= \sum_{i=1}^n p_n \sum_{\ell,m=0}^3 |\Phi_\ell\rangle\langle \Phi_m|_{MA} \otimes \sigma_\ell|\psi_i\rangle\langle \psi_i|_B \sigma_m\\ &= \sum_{\ell,m=0}^3 |\Phi_\ell\rangle\langle \Phi_m|_{MA} \otimes\left(\sum_{i=1}^n p_n\sigma_\ell|\psi_i\rangle\langle \psi_i|_B \sigma_m\right) \end{align}
So now the teleportation protocol can proceed just the same as with pure states: Perform the measurement $\{|\Phi_\ell\rangle\langle \Phi_\ell|\}$ on $MA$ with outcome $\ell$ leaving system $B$ in the state $\sigma_\ell \rho \sigma_\ell$, classically transmit the measured bit $\ell$ to $B$, and then have $B$ perform a recovery operation $\sigma_\ell$ to get back the correct state.
It might help to view how this works using an ensemble description of $\rho$ where we can describe $\rho$ as a classical distribution over states $|\psi_i\rangle$ that are drawn with probabilities $p_i$: $$\tag{5} \rho = \{(p_1, |\psi_1\rangle), \dots, (p_i, |\psi_i\rangle), \dots (p_n, |\psi_n\rangle)\} $$ From this perspective, every time you run the teleportation protocol you can imagine that you are actually randomly drawing and teleporting a pure state $|\psi_i\rangle$ with probability $p_i$. Then the the distribution of teleported states is still just $\{(|\psi_i\rangle, p_i)\}$. This makes it clear that the mixed state teleportation will work regardless of what entangled state over $AB$ you begin with.
Additionally, what would happen if $|\phi\rangle$ is one of the pure states $|\psi_i\rangle$ that $\rho$ is composed of?
I don't think this make sense. To teleport an $s$-dimensional state you will generally require entanglement over a $2s$-dimensional system $AB$. For example, it would not make sense for us to include $|\Phi_0\rangle \langle \Phi_0|$ in the ensemble $\rho$ because the dimensionalities of systems $M$ and $AB$ are off by a factor of two
$^1$ There are many choices for Bell basis states; for concreteness you can consider the basis $$ \{|\Phi_0\rangle, (\sigma_1\otimes I)|\Phi_0\rangle, i(\sigma_1\otimes \sigma_3)|\Phi_0\rangle, (I\otimes \sigma_3)|\Phi_0\rangle\} $$
What do you mean by teleportation "through" some state?
In essence, teleportation just swaps the states of stationary qubits:
$$
|\psi\rangle_A | \Phi_{start} \rangle_{AB} ~~\longrightarrow~~ |\Phi_{end}\rangle_{AA} |\psi\rangle_B
$$
Here $| \Phi_{start} \rangle$ and $|\Phi_{end}\rangle$ are some Bell states.
Two of 3 particles belong to Alice and one to Bob.
If you put some mixed state into the algorithm then the result will be the same: $$ \rho \otimes |\Phi_{start} \rangle \langle \Phi_{start} | ~~\longrightarrow~~ |\Phi_{end}\rangle \langle \Phi_{end} | \otimes \rho. $$
To understand this intuitively you can think of a mixed state $\rho$ as a classical probability distribution over states $|\psi_i\rangle$ (it doesn't matter that the choice of $\psi_i$ is not unique in general). So, the initial state of teleportation will be a mixture of $|\psi_i\rangle_A | \Phi_{start} \rangle_{AB}$ with the same weights as in $\rho$. What we'll get after Alice's measurement? Measuring a mixed state is equivalent to measuring each component with the assumption that results will coincide. So, after Alice obtained one of 4 possible results we can think that the measurement of each component $i$ gave the same result. The post-measurement state will be the mixture of post-measurement states for each component, e.g. a mixture of $|\Phi_{end}\rangle_{AA} |r_i\rangle_B$, again with weights from $\rho$. Since the results are the same for each component then Bob's final correction will also be the same. That is, the final state will be a mixture of $|\Phi_{end}\rangle_{AA} |\psi_i\rangle_B$ – with the same weights as in $\rho$.
