The constant $\alpha$ in your question can be in fact written 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
$$
g=D^nE_k-\frac{(k) _ n}{12^n}\sum_{r=0}^n(-1)^{n-r}\binom{n}{r}E_{k+2n-2r}E_2^r
$$
satisfies
$$
g\biggl(\frac{a\tau+b}{c\tau+d}\biggr)=(c\tau+d)^{k+2n}g(\tau).
$$