I have not read this entire thread, but I can put it into a mathematical form that would tell you with greater precision how much time you would have. I will not consider Peukert exponent or capacity considering the size of the battery and the normal loading nature in this analysis. This is a simple analysis.
Say we have a 1100mAh (1.1Ah) 3.6V nominal battery (4.2V to 2.6V) on a 2.5 ohm nominal coil(Resistance varies as a function of the coil temperature, but we're going to say it's 2.5ohm regardless of temperature for the sake of simplicity). From our battery we can get a more useful and commonly understood rating of Wh by multiplying the voltage times the current capacity.
3.6V*1.1Ah=3.96Wh to a safe full depth of discharge
Edit: Which is actually to 2.6V, so we're going to multiply the capacity ratio I pulled from one of my favorite batteries, the Panasonic NRC18650B
http://e-motion.lt/wp-content/uploads/2013/07/NCR18650B1S.pdf
According to the curve chart on page 2, following the 2C curve, the capacity is 1850mAh out of 3300mAh at 3.2V
1850Ah/3300Ah=0.561ul
3.96Wh*0.561ul=2.22Wh
1 Wh mean you can supply 1 Watt for one hour. Now, understand that this one hour is actual firing time. We now have to determine how many watts our coil will take. With the assumption that this is a constant voltage supply at 3.6V, this makes things much simpler in calculating our load. V^2/R=P
3.6V^2/2.5ohm = 5.184W
Now that we know our load, we can compare it to our Wh capacity. By quick observation we know the answer will be less than an hour.
2.22Wh / 5.184W = 0.4282 hours or 1542 seconds
Now remember, that is actual firing time. Say you are firing 4 seconds per vape, every minute.
1542 seconds / (4 seconds/minute) = 385.5 minutes or 6 hours 25 minutes 30 seconds
Now, this answer means that if you took one puff for 4 seconds every minute your battery will last ~6.5 hours assuming you were able to squeeze every bit of juice out of your battery until your device cut you off.