If we assume that the air in that conditions is a perfect gas, we can compute the exhaust velocity.
Let's assign the state ① to the condition before the hole and ② to the condition outside the perforation.
In the hypothesis of energy conservation from state ① to state ②,the total enthalpy per unit mass is hence conserved:
$$
h_0 = c_pT + \frac{u^2}{2}
$$
being $c_p$ the specific heat at constant pressure.
If the total enthalpy is conserved we can write:
$$
h_{01} = h_{02} \rightarrow c_pT_{01} = c_pT_2 + \frac{u_2^2}{2}
$$
$T_{01}$ is the total (or stagnation) temperature in the state ①.
We can now isolate the velocity of the gas that is flowing out from the hole $u_2$:
$$
u_2 = \sqrt{2c_p(T_{01}-T_2)} = \sqrt{2c_pT_{01}\left( 1-\frac{T_2}{T_{01}} \right)}
$$
with minor rearrangements using the equation of adiabatic for the temperature ratio:
$$
\frac{T_2}{T_{01}} = \left(\frac{p_2}{p_{01}}\right)^\frac{k-1}{k}
$$
we get the final equation for the exhaust velocity:
$$
u_2 = \sqrt{2c_pT_{01}\left(1-\left(\frac{p_2}{p_{01}}\right)^\frac{k-1}{k}\right)}
$$
if we are in space we can assume that:
$$ p_2 \simeq 0 $$
Now, we need to know what are the values of $p_{01}$ and $T_{01}$. They are so-called total quantities and are related to the speed of the gas we're considering at that state:
$$ T_{01} = T_1\left( 1 + \frac{k-1}{2}M_{01}^2 \right) $$
$$ p_{01} = p_1\left( 1 + \frac{k-1}{2}M_{01}^2 \right)^\frac{k}{k-1} $$
We can moreover assume that the velocity of the air inside the vessel is negligible, so that the Mach number $M_{01}$ is $\simeq$ 0, therefore the total pressure and total temperature are the same as the static ones.
The only unknown we have now is the static temperature inside the vessel $T_1$. To compute it we need to know at least the density of the gas inside the vessel or, equivalently, volume and mass of the air.
$$ T_1 = \frac{p_1}{\rho_1 \mathcal{R}_\text{air}}$$
Once got the exhaust velocity the thrust is proportional to it.
