In "Complex Geometry" by Huybrechts, he states the following version of the Nakano identity on page $240$:
Let $X$ be a Kähler manifold and $(E,h)$ a holomorphic hermitian vector bundle on $X$. Then if $\nabla$ is the Chern connection on $E$: \begin{align*}[\Lambda,\bar{\partial}_E]=-i(\nabla^{1,0})^*\end{align*}
In the proof he appears to say that if, over some trivialising neighbourhood $U$ of $E$ where $\nabla_E = d+A$, then $\nabla_{\check{E}} = d - A$, instead of of $d - A^T$. He also doesn't state that if $E$ is trivialised over an open set $U$, then $\bar{\partial}_E = \bar{\partial}$, which I think would simplify the proof somewhat. In fact I think we have the following:
Let $X$ be a Kähler manifold and $(E,h)$ a holomorphic hermitian vector bundle. Then if $\nabla$ is any connection on $E$: \begin{align*}[\Lambda,\bar{\partial}_E]=-i(\nabla^{1,0})^*\end{align*}
I couldn't find any other references to any other form of Nakano identity (other than the Bochner-Kodaira-Nakano formula, which I am using the Nakano identity to prove) but it seemed odd that the Chern hypothesis be used unnecessarily, whence my question:
Am I missing something, or does the stronger version of the theorem really hold?
The stronger version of Nakano's identity that you have mentioned cannot be true. You wrote
"Let X be a Kähler manifold and (E,h) a holomorphic hermitian vector bundle. Then if ∇ is any connection on E:
$[\Lambda, \bar{\partial}] = -i(\nabla^{1,0})^*$"
One can get a contradiction to this more general statement in the following way. Indeed, we know the Nakano identity holds for the Chern connection, so just cook up a new connection by adding to the Chern connection a non-trivial smooth $(1,0)$ form with values in $\operatorname{End}(E)$. Then you see that the right-hand side of the above equation gives two different answers when applied to the Chern connection and when applied to the new connection. On the other hand, the left-hand side is the same for the Chern connection and the new connection, as it only depends on the Kähler structure on $X$ and the holomorphic structure on $E$. We thus arrive at a contradiction.
Edit: I have edited my answer above in light of @JackLee's comments below. I had incorrectly written that one could cook up a different connection (than the Chern connection) by adding a smooth $(1,0)$-form to the Chern connection which is also compatible with $h$. As Jack Lee pointed out, one cannot get a new connection this way. However, for my argument to hold, I only really need to add a smooth $(1,0)$-form to the Chern connection, without requiring compatibility with $h$. One can still arrive at a contradiction this way, so the proposed generalized Nakano's identity cannot hold. I hope my answer is clearer.