Given the following Partial Trace
\begin{equation} P = \big( I \otimes \langle e_i| \big) \big[ A(B \otimes \mathcal{I} \big]^n \big( I \otimes | e_0 \rangle \big), \end{equation} where $dim(P) = 2 \times 2$, $dim(|e_j\rangle) = (2n+1) \times1$, $dim(I) = 2 \times2$, $dim(A) = (2n+1) \times(2n+1)$, $dim(B)=2\times2$, and $dim(\mathcal{I}) = 2n+1$.
My query is whether it is possible to use some tensor algebra to simplify this partial trace further?
I will assume that the expression that you are trying to simplify is $$ P = \big( I \otimes \langle e_i| \big) \big[ A(B \otimes \mathcal{I}) \big]^n \big( I \otimes | e_i \rangle \big), $$ where $A$ has size $(4n+2)\times (4n+2)$ and we are following the Einstein-summation convention.
There isn't too much you can do with this as written. However, we can write $$ A = \sum_{j,k=1}^{2} E_{jk} \otimes A_{jk} $$ where $E_{ij} = |e_i \rangle \langle e_j|$. With that, your expression becomes $$ P = \big( I \otimes \langle e_i| \big) \big[ (E_{jk} \otimes A_{jk})(B \otimes \mathcal{I}) \big]^n \big( I \otimes | e_i \rangle \big)\\ = \big( I \otimes \langle e_i| \big) \big[ (E_{jk}B \otimes A_{jk}) \big]^n \big( I \otimes | e_i \rangle \big)\\ = \big( I \otimes \langle e_i| \big) [(E_{jk}B)^n \otimes (A_{jk}) ^n] \big( I \otimes | e_i \rangle \big)\\ = (E_{jk}B)^n \cdot \langle e_i |(A_{jk}) ^n|e_i \rangle. $$ Write $E_{jk}B = xy^T$ where $x = (1,1)^T$ and $y = (B_{11} + B_{21}, B_{12} + B_{22})^T$. We can rewrite the above as $$ P = (xy^T)^n \cdot \langle e_i |(A_{jk}) ^n|e_i \rangle \\ = (y^Tx)^{n-1} \cdot \sum_{j,k}\operatorname{tr}(A_{jk}^n) \cdot xy^T. $$