The constant $\alpha$ in your question can be in fact written explicitly as $(k)_n/12^n$, where $(a)_n=\Gamma(a+n)/\Gamma(n)$ is the Pochhammer symbol (shifted factorial) and $k$ denotes the (even) weight of the corresponding Eisenstein series.

Your observation is indeed related to the Rankin--Cohen brakets; see Section 5.2 in <a href="http://dx.doi.org/10.1007/978-3-540-74119-0_1">[D. Zagier, Elliptic modular forms and their applications, *The 1-2-3 of modular forms*, Universitext (Springer, Berlin, 2008), pp. 1–-103]</a>. Preserving the notation $D$ of Zagier's lectures for your differential operator and picking a modular form $f$ of weight $k$, one can show that $D^nf$ transforms under the modular group as
$$
D^nf\biggl(\frac{a\tau+b}{c\tau+d}\biggr)
=\sum_{r=0}^n\binom{n}{r}\frac{(k+r)_{n-r}}{(2\pi i)^{n-r}}
c^{n-r}(c\tau+d)^{k+n+r}D^rf(\tau),
$$
by the induction on $n\ge 0$. In addition, the function $E_2(\tau)$ transforms as
$$
E_2\biggl(\frac{a\tau+b}{c\tau+d}\biggr)
=\frac{12c(c\tau+d)}{2\pi i}+(c\tau+d)^2E_2(\tau).
$$
Therefore, it remains to verify that the difference
$$
g_n=D^nE_k-\frac{(k) _ n}{12^n}\sum_{r=0}^n(-1)^{n-r}\binom{n}{r}E_{k+2n-2r}E_2^r
$$
satisfies
$$
g_n\biggl(\frac{a\tau+b}{c\tau+d}\biggr)=(c\tau+d)^{k+2n}g_n(\tau).
$$

**Technicalities.**
Indeed, I found the remaining details quite boring, but going through my yesterday
writing I have realised that your expectation *fails* already for $D^5E_4$ and $D^4E_6$.
Here is my explanation why.

Because $g_n(\tau)$ is a $q$-series, so it is invariant under $\tau\mapsto\tau+1$,
we can restrict to verifying the claim under the transformation $\tau\mapsto-1/\tau$
(that is, $a=d=0$, $b=-1$, and $c=1$). Then (we take $s=n-r$ in the above formula)
$$
D^nE_k(-1/\tau)
=\sum_{s=0}^n\binom ns\frac{(k+n-s) _ s}{(2\pi i)^s}\tau^{k+2n-s}D^{n-s}E_k(\tau)
$$
and
$$
\begin{aligned}
&
\frac{(k) _ n}{12^n}\sum_{r=0}^n(-1)^{n-r}\binom nrE_{k+2n-2r}E_2^r\bigg|_{\tau\mapsto-1/\tau}
\cr &\qquad
=\frac{(k) _ n}{12^n}\sum_{r=0}^n(-1)^{n-r}\binom nr\tau^{k+2n-r}E_{k+2n-2r}\biggl(\tau E_2+\frac{12}{2\pi i}\biggr)^r
\cr &\qquad
=\frac{(k) _ n}{12^n}\sum_{r=0}^n(-1)^{n-r}\binom nr\tau^{k+2n-r}E_{k+2n-2r}\sum_{s=0}^r\binom rs\frac{12^s}{(2\pi i)^s}\tau^{r-s}E_2^{r-s}
\cr &\qquad
=\frac{(k) _ n}{12^n}\sum_{s=0}^n\binom ns\tau^{r-s}\frac{12^s}{(2\pi i)^s}\tau^{k+2n-s}
\sum_{r=s}^n(-1)^{n-r}\binom{n-s}{r-s}E_{k+2n-2r}E_2^{r-s}.
\end{aligned}
$$
Subtracting the latter from the former we obtain
$$
\begin{aligned}
g_n(-1/\tau)
&=\sum_{s=0}^n\binom ns\frac{(k+n-s) _ s}{(2\pi i)^s}\tau^{k+2n-s}g_{n-s}(\tau)
\cr
&=\tau^{k+2n}g_n(\tau)+\sum_{s=1}^n\binom ns\frac{(k+n-s) _ s}{(2\pi i)^s}\tau^{k+2n-s}g_{n-s}(\tau).
\end{aligned}
$$
Therefore, $g_n(-1/\tau)=\tau^{k+2n}g_n(\tau)$, hence $g_n(\tau)$ is a modular form (of weight $k+2n$),
if and only if the additional sum over $s$ vanishes, that is, $g_{n-s}=0$ for $s=1,\dots,n$. 
The latter however does not happen when $k+2n>12$.