7. Given the parabola $x^{2}=-8 y$ whose directrix intersects the $y$-axis at point $A$, a line through $A$ intersects the parabola at points $M$ and $N$. Point $B$ lies on the axis of symmetry of the parabola, and $\left(\overrightarrow{B M}+\frac{1}{2} \overrightarrow{M N}\right) \perp \overrightarrow{M N}$. Then the range of $|\overrightarrow{O B}|$ is $\qquad$
7. $(6,+\infty)$.

As shown in Figure 3.
$$
\begin{array}{l}
\text { White }\left(\overrightarrow{B M}+\frac{\overrightarrow{M N}}{2}\right) \perp \overrightarrow{M N} \\
\Rightarrow \overrightarrow{B M} \cdot \overrightarrow{M N}=-\frac{1}{2}|\overrightarrow{M N}|^{2} \\
\Rightarrow \overrightarrow{M B} \cdot \overrightarrow{M N}=\frac{1}{2}|\overrightarrow{M N}|^{2} \\
\Rightarrow|\overrightarrow{M B}| \cos \angle B M N=\frac{1}{2}|\overrightarrow{M N}| .
\end{array}
$$

This indicates that the projection of $\overrightarrow{M B}$ in the direction of $\overrightarrow{M N}$ is exactly half the length of segment $M N$. Therefore, the projection of point $B$ on segment $M N$ is the midpoint of $M N$, denoted as $E$.

Let the line $M N$ intersect the $x$-axis at point $C$. Then $O, C, E$, and $B$ are concyclic.
Thus, $|A O||A B|=|A C||A E|$, which means
$$
2(2+|\overrightarrow{O B}|)=|A C||A E| \text {. }
$$

Therefore, we only need to find the range of $|A C||A E|$.
Let $l_{M N}: y=k x+2, C\left(-\frac{2}{k}, 0\right)$.
Substituting into the parabola equation gives
$$
x^{2}+8 k x+16=0 \text {. }
$$

Then $E\left(-4 k,-4 k^{2}+2\right)$, with $\left(k^{2}>1\right)$.
Hence (|AC||AE|)
$$
\begin{array}{l}
=\left(\frac{4}{k^{2}}+4\right)\left(16 k^{2}+16 k^{4}\right) \\
=64\left(k^{2}+1\right)^{2}>256 \\
\Rightarrow|A C||A E|>16 \\
\Rightarrow 2(2+|\overrightarrow{O B}|)>16 \\
\Rightarrow|\overrightarrow{O B}|>6 \\
\Rightarrow|\overrightarrow{O B}| \text { is in the range }(6,+\infty) .
\end{array}
$$