Suppose we have a partially separable 3-qubit state
$$φ = \left(a_0\left|0\right\rangle + a_1\left|1\right \rangle\right) \otimes \left(b_{00}\left|00\right \rangle + b_{01}\left|01\right \rangle + b_{10}\left|10\right \rangle + b_{11}\left|11\right \rangle\right)$$
i.e. the second and third qubits are entangled with each other, but the first qubit is separable.
The unseparated state is given by
$$φ = c_{000}\left|000\right\rangle + c_{001}\left|001\right\rangle + c_{010}\left|010\right\rangle + c_{100}\left|100\right\rangle+ c_{011}\left|011\right\rangle + c_{101}\left|101\right\rangle + c_{110}\left|110\right\rangle + c_{111}\left|111\right\rangle $$
with $c_{ijk} = a_i b_{jk}$.
Now suppose I want to apply a unitary transformation to the first two qubits, given by a 4x4 unitary matrix:
$U = (u_{nm})$
Q: What does the overall 8x8 matrix look like which will also calculate the effect on the third qubit (which won't leave the interaction unchanged)?
The entanglement of the qubits does not matter. All you need to do is tensor $U$ with $\begin{pmatrix}1&0 \\ 0 & 1\end{pmatrix}$, which is the effect of $U$ on the third qubit. You'll get $U' = \begin{pmatrix}U&0 \\ 0 & U\end{pmatrix}$ or a permutation thereof.
As a linear transformation, $U' = U \otimes \text{id}$, so you'd have $$U' \left|xyz\right\rangle = U'(\left|xy\right\rangle \otimes\left|z\right\rangle) = U\left|xy\right\rangle \otimes \left|z\right\rangle.$$