I don’t know if this is proved anywhere. It looks like one can obtain a formula for the Schur functions in terms of immanants of replicated matrices. Let me introduce the following notation. Define a type $T$ as an ordered set $(i_1,\dots, i_n)$ such that $\sum_k i_k=n$. This defines a partition $\nu(T)$ obtained by rearranging the $i_k$ in decreasing order. For a type $T$, define $A_T$ as the matrix $A$ with the first row and column repeated $i_1$ times, second row and column repeated $i_2$ times etc. This is still an $n\times n$ matrix. Then we can obtain
\begin{equation}
s_\lambda(\mu_1\dots,\mu_n)=\sum_T d_\lambda(A_T)\,,
\end{equation}
where $\mu_i$ are the eigenvalues of $A$, $m_\lambda$ is the dimension of the symmetric group irrep $\lambda$ and the sum is over all types $T$ such that the partitions $\nu(T)$ have a non-zero Kostka number $K_{\nu,\lambda}\neq 0$.

To obtain this, we can use Schur-Weyl duality. First, write the immanant as
\begin{equation}
d_\lambda(A) = \sum_\sigma \chi_\lambda(\sigma) \langle 1,\dots,n| A^{\otimes n} |\sigma(1),\dots,\sigma(n)\rangle
\end{equation}
This is equal to
\begin{equation}
d_\lambda(A) = \frac{n!}{m_\lambda}\langle 1,\dots,n| A^{\otimes n}\Pi_\lambda | 1,\dots,n\rangle\,,
\end{equation}
where $\Pi_\lambda$ is the projector on to the isotypic space of the symmetric group irrep $\lambda$. This can now be written as
\begin{equation}
d_\lambda(A) = \frac{n!}{m_\lambda} \text{Tr}(A^{\otimes n}\Pi_\lambda |1,\dots,n\rangle\langle 1,\dots,n|) \,.
\end{equation}
Finally, this can be written as
\begin{equation}
d_\lambda(A) = \frac{1}{m_\lambda}\text{Tr}(A^{\otimes n}\Pi_\lambda \Pi_{1^n}) \,,
\end{equation}
where $\Pi_{1^n}$ is the projector onto the induced representation from the trivial representation of the Young subgroup corresponding to the partition $1^n$. This induced representation is sometimes called the permutation module of the partition $1^n$. The multiplicity of the symmetric group irrep $\lambda$ in this module is the Kostka number $K_{1^n,\lambda}$. In the last step, we have also used the fact that
\begin{equation}
\frac{1}{n!}\sum_\sigma \sigma |1,\dots,n\rangle\langle 1,\dots,n |\sigma^{-1} = \Pi_{1^n}\,.
\end{equation}
Let us also block diagonalize $A^{\otimes n}$. We then obtain
\begin{equation}
d_\lambda(A) = \frac{1}{m_\lambda}\text{Tr}((I_\lambda\otimes A_\lambda)\Pi_\lambda \Pi_{1^n}) \,,
\end{equation}
where $I_\lambda$ is the identity on the symmetric group irrep space and $A_\lambda$ is the part in the $GL_n$ irrep space.

Now it is clear what to do. If we add all the immanants for different types, we get
\begin{equation}
\sum_T d_\lambda(A_T) = \sum_T\frac{1}{m_\lambda}\text{Tr}((I_\lambda\otimes A_\lambda)\Pi_\lambda \Pi_{\nu(T)})= s_\lambda(\mu_1,\dots,\mu_n)\,.
\end{equation}
For the determinant, the sum over the other partitions do not matter since their projections onto the alternating irrep are zero. So we obtain that it is the Schur function in that case.