There are two stages to the flow here. During the first stage, there is some pressure loss (due to viscosity a.k.a. friction) and the air comes out the tube at some lower pressure than it went in but at the same speed (neglecting changes in density). During the second stage, the remaining pressure causes the air to come out at some speed, according to Bernoulli. These two things are coupled: How much pressure is lost depends on how fast the air is flowing, but how fast the air is flowing depends on the pressure at the exit which depends on how much pressure was lost. So you get a system of equations: Two unknowns (speed and pressure at the tube exit) and two equations (one for viscous pressure loss, then the Bernoulli equation at the exit, both of which connect pressure and speed).
If you assume incompressible laminar flow (that is the “back of the napkin” part… Turbulent and/or compressible flow would give a more accurate answer but would be more complicated), the head loss (in units of pressure) is 32VLν/D² where V is speed, L is pipe length, D is pipe diameter, and ν is kinematic viscosity.
Now, the speed is determined by the difference between Pexit and Patmosphere. According to Bernoulli, Pexit minus Patmosphere is (1/2)ρV², right? (…which is how pitot tubes give us indicated airspeed).
This means that so far we have:
Pcontainer - Pexit = 32VLν/D²
Pexit - Patmosphere = (1/2)ρV²
The unknowns are Pexit and V.
I’m going to convert this all to metric because imperial units are an abomination, a curse from the devil, so I avoid them at all costs unless work makes me use them.
172369 Pa – Pexit = 32 * V * 1.8288m * .0000148 m²/s / (0.003175m)²
Pexit – 101325 Pa = 0.5 * 1.225 kg/m³ * V²
Rearrange the first one,
172369 Pa – 32 * V * 1.8288m * .0000148 m²/s / (0.003175m)² = Pexit
[172369 Pa – 32* V * 1.8288m * .0000148 m²/s / (0.003175m)²] – 101325 Pa = 0.5 * 1.225 kg/m³ * V²
For Wolfram Alpha (because I'm starting to get lazy): (172369-32*y*1.8288*.0000148/(0.003175*0.003175))-101325=0.5*1.225*y^2
y = 277.582 m/s … which is like 80% the speed of sound, again showing that the laminar incompressible flow equations are probably not appropriate for this much pressure.
In any case, if the ID is 1/8”, then the area is 0.000007917304361m², which times 277.6 m/s gives 0.0021977m³/s which is about 4.66 cubic feet per minute.
But, again, this will not be very accurate at all due to neglecting compressibility and turbulence.
And someone please check my math because I could have easily made a mistake somewhere. But I'm within a factor of 1.5 of Greg Niehues' estimate, which is reassuring.
 Where did I get “32LVν/D²”? Googling the equations for head loss in pipe flow will return many sets of lecture notes such as this one
Re = DVρ/ν
ΔP=2*(16ν/DVρ)(L/D)(ρV²) = 32νLρV²/DVρD = 32νLV/D²
If you want, you can go into those lecture notes and use the turbulent flow equations. They involve taking out the "16/Re" and replacing it by the Colebrook or the Zigrang-Sylvester Equation, which include a roughness factor ε. (The last page of the PDF includes a table of ε values to plug in). I think this is enough fun for me for today
PS: Do you know what "nerd sniping"