The flow rate through the holes depends on the velocity, which depends on the pressure.
It's reasonable to assume that the flow inside the bag is slow enough that the pressure is hydrostatic, so $P+\rho g z=P_b$, and outside the pressure is also hydrostatic so $P+\rho g z=P_a$. The $\rho g z$ terms cancel from here, if we use gauge pressure (so set $P_a=0$) for convenience.
The pressure in a straight free jet (the rough structure you'd expect from the holes) is the same as the fluid it's entering, so $P_a$, which we've set to zero. Between the inside and the holes there's a streamline and little viscous loss, so $P_b-0=\frac{1}{2}\rho V^2$, so $V=\sqrt{2P_b/\rho}$
The volumetric flow rate (in $m^3/s$) is just $A_hV$, where $A_h$ is the total area of all holes.
To calculate the pressure: The pressure needs to be high enough to hold the weight of the bag in its inflated position. Imagine if the bag is roughly a half sphere with a flat bottom, and think of a horizontal slice just above the bottom: about half the mass of the bag is above, and the pressure is acting on a circular slice, so the force from the pressure is $P_b \pi r^2$, where $r$ is the radius of the inflated bag.
This pressure has to equal the weight of the half of the bag material, so $P_b \pi r^2=m_bg/2$ where $m_b$ is the mass of the bag
So the flow rate is $A_h\sqrt{m_bg/\rho\pi r^2}$
This isn't exact, because the final shape of the bag probably won't be half-spherical, but should be the right order of magnitude - probably double it to be on the safe side
Adding extra flow will increase the pressure, which gets reacted by tension in the bag walls, and makes the bag closer to a full sphere
