Determinant of determinant is determinant? Looking at this question, I am thinking to consider the map $R\to M_n(R)$ where $R$ is a ring, sending $r\in R$ to $rI_n\in M_n(R).$ Then this induces a map. $$f:M_n(R)\rightarrow M_n(M_n(R))$$ 
Then we consider another map $g:M_n(M_n(R))\rightarrow M_{n^2}(R)$ sending, e.g. $$\begin{pmatrix}
\begin{pmatrix}1&0\\0&1\end{pmatrix}&\begin{pmatrix}2&1\\3&0\end{pmatrix}\\ \begin{pmatrix} 0&0\\0&0 \end{pmatrix}&\begin{pmatrix} 2&3\\5&2\end{pmatrix} \end{pmatrix}$$ to $$\begin{pmatrix}1&0&2&1\\
0&1&3&0\\
0&0&2&3\\
0&0&5&2\end{pmatrix}.$$

Is it true that $$\det_{M_n(R)}(\det_{M_n(M_n(R))}A)=\det_{M_{n^2}(R)}g(A)$$ for some properly-defined determinant on $M_n(M_n(R))?$

If this is true, then $\det_{M_n(R)}\operatorname{ch}_{A}(B)=\det_{M_n(R)}\circ\det_{M_n(M_n(R))}(f(A)-B\cdot I_{M_n(M_n(R))})=\det_{M_{n^2}}\circ g(f(A)-B\cdot I_{M_{n^2}(R)}),$ which is what OP of the linked question wants to prove.
Any hint or reference is greatly appreciated, thanks in advance.  

P.S. @user1551 pointed out that determinant is defined on commutative rings only and $M_n(R)$ is in general a non-commutative ring. So I am thinking maybe we could use the Dieudonné determinant. In any case, I changed the question accordingly.

(Edit: the OP has modified their question; this answer no longer applies.)
Your question is not well posed because determinant is defined on commutative rings only, but $M_n(R)$ in general is not a commutative ring. But there is indeed something similar to what you ask. See


*

*M. H. Ingraham, A note on determinants, Bull. Amer. Math. Soc., vol. 43, no. 8 (1937), pp.579-580, or

*John Silvester, Determinants of Block Matrices, theorem 1.


Briefly speaking, suppose $B\in M_m(R)$, where $R$ is commutative subring $R\subseteq M_n(F)$ for some field $F$ (note: I follow Silvester's notation here; his $R$ is not your $R$). If you "de-partition" $B$, you get a matrix $A\in M_{mn}(F)$. In other words, the entries of $A$ are taken from a field $F$, and if you partition $A$ into a block matrix $B$, all blocks commute.
Now Silvester's result says that $\det_F A=\det_F(\det_R B)$. Put it another way, if you take the determinant of $B$, the result is a "scalar" in $R$, which is by itself an $n\times n$ matrix over $F$. If you further take the determinant of this resulting matrix, you get a scalar value in $F$. As shown by Silvester, this value must be equal to the determinant of $A$.
Silvester's proof still applies if $F$ is an integral domain instead. I don't know how far the assumption on $F$ can be weakened to.
