Solution:

(Ⅰ) Since $a^2=4$, $b^2=1$, then $c= \sqrt{a^2-b^2}= \sqrt{3}$, thus $F_1(-\sqrt{3},0)$, $F_2=(\sqrt{3},0)$.

Let $Q(x,y)$, $P(x_0,y_0)$.

Since the moving point $Q$ satisfies $\overrightarrow{OQ} = \overrightarrow{PF_1} + \overrightarrow{PF_2}$,

then $\begin{cases} x=-\sqrt{3}-x_0+\sqrt{3}-x_0 \\ y=-y_0-y_0\end{cases}$, solving gives $x_0=-\frac{x}{2}$, $y_0=-\frac{y}{2}$,

Substituting into the ellipse equation yields: $\frac{x^2}{16} + \frac{y^2}{4} = 1$,

Therefore, the trajectory equation of the moving point $Q$ is: $\boxed{\frac{x^2}{16} + \frac{y^2}{4} = 1}$.

(Ⅱ) According to the problem, the slope of line $l$ exists,

Let the equation of line $l$ be $y=kx-2$. Let $B(x_1,y_1)$, $C(x_2,y_2)$.

Combining $\begin{cases} y=kx-2 \\ \frac{x^2}{4}+y^2=1\end{cases}$, it becomes: $(1+4k^2)x^2-16kx+12=0$,

Since $\triangle > 0$, we get $k^2 > \frac{3}{4}$. Thus $x_1+x_2= \frac{16k}{1+4k^2}$, $x_1x_2= \frac{12}{1+4k^2}$.

$S_{\triangle OBC}=S_{\triangle OAC}-S_{\triangle OAB}= \frac{1}{2}|OA|(|x_2|-|x_1|)=|x_2-x_1|$

$= \sqrt{(x_1+x_2)^2-4x_1x_2}= \sqrt{\frac{256k^2}{(1+4k^2)^2}- \frac{48}{1+4k^2}}= \frac{4\sqrt{4k^2-3}}{1+4k^2}$.

Let $\sqrt{4k^2-3}=t > 0$, it becomes $4k^2=t^2+3$. Thus $S_{\triangle OBC}= \frac{4t}{t^2+4}= \frac{4}{t+ \frac{4}{t}}\leqslant \frac{4}{2\sqrt{t\cdot \frac{4}{t}}}=1$,

Equality holds if and only if $t=2$, at this time $k=\pm \frac{\sqrt{7}}{2}$.

Therefore, $\boxed{(S_{\triangle OBC})_{\text{max}}=1}$.