Hint: You have an error in your computations. In particular in the grand canonical ensemble,
\begin{align}
\langle E \rangle \neq -\frac{\partial \log Q}{\partial \beta}.
\end{align}
Moreover, I just did the whole computation having corrected this error in the appropriate way, and it worked out the way it should.
Addendum, 2019-02-02. Details Beyond the Hint
Step 1. Recall the following definitions of the grand canonical partition function $Q$, the ensemble average energy $\langle E\rangle$ in the grand canonical ensemble, and the ensemble average particle number $\langle N\rangle$ in the grand canonical ensemble. All sums are over states $i$ of the system:
\begin{align}
Q \equiv \sum_ie^{-\beta(E_i - \mu N_i)}, \qquad \langle E\rangle \equiv \sum_i \frac{e^{-\beta(E_i - \mu N_i)}}{Q}E_i, \qquad \langle N\rangle \equiv \sum_i \frac{e^{-\beta(E_i - \mu N_i)}}{Q}N_i
\end{align}
Step 2. Show that the following identity follows from the definitions in Step 1:
\begin{align}
\langle E\rangle = -\frac{\partial \ln Q}{\partial \beta} + \mu\langle N\rangle.
\end{align}
Step 3. Show that if we take
\begin{align}
Q = V\frac{e^{\beta\mu}}{\lambda^3},
\end{align}
then
\begin{align}
-\frac{\partial \ln Q}{\partial \beta} = \frac{3}{2}\frac{\langle N\rangle}{\beta} - \mu\langle N\rangle.
\end{align}
Step 4. Combine steps 2 and 3 to obtain the desired result.
What is wrong is using a chain rule without a careful analysis of the functional dependence of all the quantities.
Let's look at your formulae. Once one gets $(\partial{\left<N\right>}/\partial{\mu})_{T,V}$, $\left<N\right>$ should be understood as a function of $(T,V,\mu)$. In order to connect this derivative to a derivative at fixed $\left<N\right>$, the best way to avoid confusion is to start from the differential of $\left<N\right>$:
$$
d\left<N\right> = \left(\frac{\partial{\left<N\right>}}{\partial{\mu}}\right)_{T,V}d\mu +
\left(\frac{\partial{\left<N\right>}}{\partial{V}}\right)_{T,\mu}dV +
\left(\frac{\partial{\left<N\right>}}{\partial{T}}\right)_{\mu,V}dT.
$$
So we get:
$$
\left(\frac{\partial{\left<N\right>}}{\partial{\mu}}\right)_{T,V}d\mu +
\left(\frac{\partial{\left<N\right>}}{\partial{V}}\right)_{T,\mu}dV = 0,
$$
valid for any finite variation of $\mu$ and $V$ at fixed $\left<N\right>$ and $T$. Thus, we can get easily
$$
\left(\frac{\partial{V}}{\partial{\mu}}\right)_{T,\left<N\right>}
= -\left(\frac{\partial{\left<N\right>}}{\partial{\mu}}\right)_{T,V}
\left(\frac{\partial{V}}{\partial{\left<N\right>}}\right)_{\mu,T},
$$
which looks like your "chain rule", but has the right sign.
A different way to get the same result follows a more "thermodynamic" path. Let's start with the inverse of the quantity you were manipulating, i.e. $(\partial{\mu}/\partial{N})_{V,T}$. By using the Gibbs-Duhem relation:
$$
d\mu = -\frac{S}{N}dT + \frac{V}{N}dP,
$$
at fixed T we have:
$$
\left(\frac{S}{N}\right)_{V,T}= \frac{V}{N}\left(\frac{\partial{P}}{\partial{N}}\right)_{V,T}=
-\frac{V}{N}\left(\frac{\partial{\mu}}{\partial{V}}\right)_{N,T}= -\left(\frac{V}{N}\right)^2\left(\frac{\partial{P}}{\partial{V}}\right)_{V,T}
$$
(Maxwell relations have been used twice to go from the second to the third and from the third to the forth formula) from which one can easily extract the relation with the isothermal compressibility.
Best Answer
First, using the partition function $Z=\sum_n e^{-\beta E_n}$, you have $$ \langle E\rangle=\frac{1}{Z}\partial_{-\beta}Z, $$$$ \langle E^2\rangle=\frac{1}{Z}\partial_{-\beta}^2Z, $$$$ \langle E^3\rangle=\frac{1}{Z}\partial_{-\beta}^3Z. $$ Second, you can show $$ \langle E\rangle=\partial_{-\beta} \ln Z, $$$$ \langle (\Delta E)^2\rangle=\langle (E-\langle E\rangle)^2\rangle=\partial_{-\beta}^2 \ln Z, $$$$ \langle (\Delta E)^3\rangle=\langle (E-\langle E\rangle)^3\rangle=\partial_{-\beta}^3 \ln Z. $$ Well, it just indicates that $\langle E^n\rangle$'s are disconnected correlation functions, while $\langle (\Delta E)^n\rangle$'s connected. And at last: $$ \langle (\Delta E)^3\rangle=\partial_{-\beta}^2 U=(k_B T^2\partial_T)^2 E=k_B^2 T^2\partial_T(T^2 C_V)=k_B^2(T^4\partial_T C_V+2T^3C_V). $$ Aha, trumpets!