$(1)$ Because the focus of the parabola is $F(0,1)$, we have $\frac{p}{2} = 1$, which implies $p = 2$.

$(2)$ From $(1)$, we know the equation of the parabola $C$ is $x^2 = 4y$. Let $A(x_1, y_1)$ and $B(x_2, y_2)$. Let the equation of line $L$ be $y = kx + 1$. Substituting into $x^2 = 4y$, we obtain $x^2 - 4kx - 4 = 0$.

As a result, $x_1 + x_2 = 4k$ and $x_1x_2 = -4$. The sum $y_1 + y_2 = k(x_1 + x_2) + 2 = 4k^2 + 2$.

Consequently, the midpoint of $AB$ is $Q(2k, 2k^2 + 1)$.

Therefore, $\left|AB\right| = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$, which simplifies as follows, knowing that $AB$ is a diameter of the circle intersecting the x-axis:

$\left|AB\right| = \sqrt{y_1 + y_2 + 2} = \sqrt{k(x_1 + x_2) + 4} = \sqrt{4k^2 + 4} = 2\sqrt{k^2 + 1}$.

In the isosceles triangle $QMN$ with $QN = NM$, angle $\angle QMN$ is acute, and using the sine rule, we have:

$\sin \angle QMN = \frac{QO}{QN} = \frac{y_Q}{\frac{1}{2}AB} = \frac{2k^2 + 1}{2\sqrt{k^2 + 1}}$.

We want to minimize $\sin \angle QMN$, and as $\sin \angle QMN$ increases as the numerator grows while the denominator stays constant, the minimum value occurs when $k = 0$ because $\sin \angle QMN$ approaches its minimum when approaching $\sin \frac{\pi}{6} = \frac{1}{2}$.

So, the minimum value of $\angle QMN$ is $\boxed{\frac{\pi}{6}}$.

At this moment, $k = 0$, and therefore, the equation of line $l$ is $\boxed{y = 1}$.