When do the Nakano identities hold? 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?

 A: 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.
A: The answer is tentative. I feel that there might be a mistake in it (see the end of the answer)
I am stuck at the same place in the textbook too, but I think it might just be a typo in the proof, instead of some more serious work going on in the background. Let's try just plowing through some computations.
As you suggested, $\nabla_{E^{\vee}}=d-A^t=d+\bar{A}$ (where we used $A^*=-A$ from the first condition that Chern connection is hermitian). Then $$(\nabla_{E}^{1,0})^*=-\bar{*} \circ (\partial+ (\bar{A})^{1,0})\circ \bar{*}=-\bar{*} \circ (\partial+ (\overline{A^{0,1}}))\circ \bar{*}=\partial^* +(\overline{A^{0,1}})^*$$
Second Chern condition of being compatible with holomorphic structure indeed tells us that $$\overline{\partial_E}=\overline{\partial}+A^{0,1}$$and looks a bit weird since we are not in the holomorphic trivialization.
Then we compute as in the book: $$[\Lambda, \overline{\partial_E}]+i (\nabla_{E}^{1,0})^*=[\Lambda, \overline{\partial}]+i\partial^*+[\Lambda, A^{0,1}]+i(\overline{A^{0,1}})^*=[\Lambda, A^{0,1}]+i(\overline{A^{0,1}})^*$$where for the first two summands we used the usual Kahler identity. The last expression is linear and we use the trick from remark 4.2.5. to show that it in fact vanishes pointwise.
PS. In fact, it indeed seems that the first condition (hermitian connection) doesn't affect the proof at all - there will be some weird adjoint in the last term and that's all.
