I assume your "mostly sleeping" circuit consumes little average power, and you intend the li-ion cell to discharge slowly (tens of hours or more), i.e., the discharge rate would be somewhere below C/20.
Now, while you can and do discharge the cells down to 2.5V (or 2.8V or whatever), this is done in high-current applications, because the ESR of the cell is fairly high near zero State-of-Charge - especially in cold environment.
For you, such an issue isn't true - this means, there is little resistive drop over the Open-Circuit Voltage of the cell. Now, for a typical li-ion cell, the OCV of an empty cell is around...
3.4 V!
So, if you employ an almost zero-dropout LDO, the chances are, you are already extracting 100% (or very close) of the charge, and won't need to run down to 2.5V. Running down there wouldn't give you much extra energy, and would risk damaging the cell. I mean, if you read the cell datasheets carefully, they tend to specify the end-of-discharge condition like this:
2.5V @ 0.5C discharge current.
If you are not using this exact current, you are exceeding the specification, and going below their definition of 0% State-of-Charge. This may sound unintuitive, since you are using lower current than specified.
I.e., what nctnico said is exactly true for low-current applications.
For really high-power stuff (running near the maximum ratings of the cells), especially in cold weather, there can easily be 20-30%, even closer to 50% charge left at 3.3V.