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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 (1,0) form with values in End(E) (or with values in the bundle of skew-hermitian endomorphisms of E with respect to $h$, if you insist on the connection being compatible with $h$). 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. I hope this helps.

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: $$[Lambda, overline{partial_E}]+i (nabla_{E}^{1,0})^*=[Lambda, overline{partial}]+ipartial^*+[Lambda, A^{0,1}]+i(overline{A^{0,1}})^*=[Lambda, A^{0,1}]+i(overline{A^{0,1}})^*$$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.