To calculate L293D power usage, you need to lookup the datasheet and check the Voh and Vol. These are the High level voltage and Low level voltage.
On Texas Instruments datasheet, we have:
So you will lose about 1.4V when your output is -0.6A, and 1.2V when your output is 0.6A. If your motor uses 2 outputs (bi-directional) then you have to add the losses.
Then, to get the power dissipated on the power side of the L293D, you take this voltage drop and multiply by the motor´s current.
Lets say your motor is using 0.6Amps and your power source is 5V + 24V @ 25°C (just like in the datasheet).
The power loss in the L293D is aprox.: (1.4V + 1.2V) x 0.6A = 1,56W.
Since you´ve lost 2.6V in the L293D, your motor will use: (24V - 2.6V) x 0.6A = 12,84W.
To measure this directly on your real circuit, use a multimeter in Ampere mode to measure current, and Volt mode to measure the voltage on the motor terminals when it is on.
As for the arduino, your battery has 2000mAh and I assume it is a 7,4V battery. So it has 2000mAh x 7,4V = 14800mWh of energy stored in it.
If you are running your arduino at 5V, then: 14800mWh / 5V = 2960mAh at 5V.
Now let´s say your arduino and it´s modules, the L293D´s and the motors are using 500mA average, then your battery will last: 2960mAh / 500mA = 5.92 hours in theory.
Why "in theory"? because in real life it is impossible to extract all the energy from inside a battery. A minimum ammount of energy has to remain inside it and some power is lost when you extract it´s energy.
Of course, this is a very raw calculation. In reality, you have to calculate power used in every component. But by calculating the "big guns" first you can get an idea of where you stand.
Basic power loss calculation is done by multiplying voltage drop by the current in each circuit. Voltage drop is the difference between your volts before and after the component.