This problem can be solved using Bernoulli's principle for fluid dynamics. We consider a streamline from a point far outside the container (point 1) to a point just inside the hole (point 2). We can assume the air is an incompressible fluid for this initial moment and that the change in gravitational potential energy is negligible ($h_1 = h_2$).

Bernoulli's equation is:

$$p_1 + \frac{1}{2}\rho v_1^2 = p_2 + \frac{1}{2}\rho v_2^2$$

At point 1 (far outside), the pressure is atmospheric pressure, $p_1 = p_0$, and the air speed is negligible, $v_1 = 0$. At point 2 (just inside the hole), the pressure is the internal container pressure, $p_2 = p$. We want to find the speed $v_2 = v$.

Substituting these values into Bernoulli's equation:

$$p_0 + \frac{1}{2}\rho (0)^2 = p + \frac{1}{2}\rho v^2$$ $$p_0 = p + \frac{1}{2}\rho v^2$$

Solving for the speed $v$:

$$\frac{1}{2}\rho v^2 = p_0 - p$$ $$v^2 = \frac{2(p_0 - p)}{\rho}$$ $$v = \sqrt{\frac{2(p_0 - p)}{\rho}}$$

Since the problem states that $p \ll p_0$, we can approximate $p_0 - p \approx p_0$.

$$v \approx \sqrt{\frac{2p_0}{\rho}}$$

However, the more precise answer includes $p$.