A simple silicon diode will drop 0.5 - 0.7V (probably closer to 0.7 at 10mA). So, if you just need "around" 2.5V unregulated, a diode may do the trick. If you'd rather err on the high-side (when the battery is new & fully charged) a Schottky diode will drop about 0.3V.
There are a couple of issues with regulators... The battery voltage is going to drop as the battery ages. You can look for a low dropout (LDO) regulator, but when the battery drops down near 2.5V, at some point the regulator will drop out of regulation.
There are boost-buck (switching) regulators that can either reduce or boost the voltage and they can be very efficient. But a low power, I don't think you'll get anywhere the rated efficiency and the battery will probably be drained quickly.
Any regulator circuit requires some power just to operate, so just hooking-up a regulator is going to drain the tiny battery to some extent.
Well. Actually I'll run the Atmega328(@8Mhz) with this 2.5v.
I'm not sure If I can connect the 3.3 battery directly to the arduino.
My total current is about 8mA at peak . my project is a variometer that emites beeps(it may cause the voltage to drop if i don't use a regulator and Atmega may reset or something. Am I right)?
Or I can use a the 3v battery cell directly to the Atmega?