Deciding between 7.4V and 11.1V power
Work at the lowest possible voltage. If you draw \$100mA\$ at \$7.4V\$ that's \$740mW\$, at \$11.1V\$ that's \$1100mW\$, or 50% more. The capacity of the \$11.1V\$ is also 50% higher, so there's no difference there, apart from the \$11.1V\$ battery being bulkier and heavier and more expensive.  

But if the load is resistive there will be indeed a difference. Suppose the resistance is \$74\Omega\$, then the power drawn from the 7.4V battery is  


  \$P = \dfrac{V^2}{R} = \dfrac{(7.4V)^2}{74\Omega} = 740mW\$  


while the power drawn from an 11.1V battery is  


  \$P = \dfrac{V^2}{R} = \dfrac{(11.1V)^2}{74\Omega} = 1665mW\$  


or 2.25 times as much. Since the \$11.1V\$ battery's capacity is only 1.5 times as high it will drain faster.

edit
Kevin wants to read about linear and maybe other regulators
If the battery feeds a regulator then the circuit's power will be independent of the battery's voltage. Say you regulate to 5V and draw 100mA from that, then the circuit will always consume 500mW, no matter what the battery's voltage is, because the 5V is fixed. If it's a linear regulator then the voltage difference between battery and regulator output will cause dissipation in the regulator. For the 7.4V battery:  


  \$P = \Delta V \cdot I = (7.4V - 5V) \cdot 100mA = 240mW\$  


If the battery's voltage is the 11.1V however this becomes  


  \$P = \Delta V \cdot I = (11.1V - 5V) \cdot 100mA = 610mW\$  


so your power loss in the regulator will be higher for the higher voltage battery. But just like in the first example the battery's longevity will be the same: 


  Battery life = \$ \dfrac{1000mAh}{100mA} = 10h \$


since the 11.1V battery has a higher capacity. It's just a pity to waste that capacity instead of using it to get a 50% longer battery life.
A switcher (SMPS, or Switch-Mode Power Supply) solves this. A switcher allows high efficiency power conversion. Let's assume an ideal switcher with a 100% efficiency. If you draw 100mA from the 5V output this will represent only 68mA from the 7.4V battery, or 45mA from the 11.1V battery. That's because the power remains the same:  


  \$P = V \cdot I = 5V \cdot 100mA = 7.4V \cdot 68mA = 11.1V \cdot 45mA \$


A 1000mAh battery will then last  


  \$ \dfrac{1000mAh}{68mA} = 14.8h (7.4V) \$ or \$ \dfrac{1000mAh}{45mA} = 22.2h (11.1V) \$


so here we see that we have a 50% longer battery life for the bigger battery, since we don't have conversion losses. Remember that we calculated for a 100% efficient switcher. We could recalculate for a more realistic 85% conversion efficiency, and find the same ratio. I'll leave that as an exercise for the interested reader :-).