Your 8.75mA is high by a factor of three. You've effectively summed the charge transferred in a 3 second period and not divided by 3s in order to get coulombs-per-second, or amps.
So the average current is about 2.88mA and theoretically speaking, you could perhaps get (2000mAh/2.88mA)=693hr=28 days from the batteries. However, the problem is that batteries lose their capacity more-rapidly at higher drain currents than at lower drain currents. If you're using a very high-quality, high-current battery (e.g. an Eneloop NiMH) then you probably will get the rated capacity from it.
A cheap battery under 200mA+ peak load will discharge faster than indicated by its rated capacity, which means that the actual capacity available to you will be reduced. However according to that Toshiba datasheet, the discharge curve is pretty linear even down around 3-5 ohms load (where you are with your peak currents), so the performance calculation above is probably close to valid if you use the Toshiba cells. For a cheap cell or even worse, Zinc battery, it will be much much worse.
The second issue is that the rated capacity is for discharge to a fairly low voltage (0.9V), which may be a lower voltage than can support your circuitry. If your circuit fails at 4.5V (1.1V/cell) but the battery rating was computed for a discharge down to 0.9V, then you will clearly get less usable capacity than the rated capacity.
After 18 days, you will have used 18/28=65% of the capacity. If you take the 2 ohm 140-minute voltage discharge graph at bottom left as being representative of the cell's discharge-time/voltage curve (it may or may not be, because it's a very heavy discharge and you might have a constant-current instead of constant-resistance load) and look at it at 0.65*140=90 minutes, that gives you a voltage of about 1.05V/cell. So you might expect to see about 4.2V on your 4-cell pack after 18 days of use. There are lots of uncertainties there though.
