Can we calculate the "previous hash value" in the SHA256 compression function with a known message block and final hash?
Can we compute the intermediate hash $H_n$ using just the fully known final message block, and the final hash output $H_{n+1}$?

No, that's computationally infeasible. The only known way to find $H_n$ is when we know all (or enough of) the message, which allows to directly compute $H_n$ in the forward direction (or use brute force search of a small unknown fraction of the message, validating a guess by checking if the final $H_{n+1}$ matches).
The reason we can not compute $H_n$ is that the Davies-Meyer round function $(H_n,M_n)\mapsto H_{n+1}$ restricted to fixed known $M_n$ is computationally difficult to invert.
That follows from its construction as $H_{n+1}\gets E(\text{key}\gets M_n, \text{block}\gets H_n)\boxplus H_n$ where $E$ is a public block cipher and $\boxplus$ is a group operation¹ on $\{0,1\}^{256}$ (256-bit bitstrings).
If some fairy told us the output $X$ of $E(\text{key}\gets M_n, \text{block}\gets H_n)$, and since we know $M_n$ and $H_{n+1}$, we could find $H_n$ in two ways:

by using the decryption function $D$ matching $E$, with $H_n\gets D(\text{key}\gets M_n, \text{block}\gets X)$
by inverting $X\boxplus H_n\mapsto H_{n+1}$ to find $H_n$ from the known $H_{n+1}$ and the $X$ given by the fairy.

But we have no fairy, and there is no known way out of this chicken-and-egg problem. That's even provable under a model of $E$ as an ideal cipher. That's by design of the Davies-Meyer round function.
Also: it is likely that for a sizable fraction of the final hash $H_{n+1}$ and message size (equivalently, values of $M_n$ ), the known relation $H_{n+1}=E(\text{key}\gets M_n, \text{block}\gets H_n)\boxplus H_n$ leaves several possible values for $H_n$.

¹ $\boxplus$ happens to be addition of 256-bit bitstrings ignoring carry bit between blocks of 32 bits.