Consider the operation of this circuit. When the switch (SW) is "on," (tied to ground) the inductor has Vin across it, so the current in the inductor is increasing. Since V = L*di/dt\$V = L\times di/dt\$, you can calculate the rate of change of current for whichever value of inductor you choose. Since the frequency is fixed, and the maximum duty cycle is 90%, you can also calculate the maximum amount that the current can increase on each cycle. So, a larger inductor value means that the circuit may take more cycles to adjust to a sudden change in load (poorer load regulation) and may have a longer start-up.
When the switch is open, (assuming that the power supply is working) the energy in the coil is 0.5 * L * i(peak)^2\$0.5 \times L \times {i_{peak}}^2\$. This is how much energy you could theoretically get from each cycle (your worst case), so a larger inductance value will result in lower peak current from your battery for a given load. Multiply this energy (joules) by the number of cycles per second, and you get the power in watts. Your inductor current rating should be this calculated peak current, which will be different for different inductor values. In practice, your circuit will not charge/discharge the inductor completely on every cycle, but this will be a safe value.
Power loss in the inductor is related to the DCR so lower is better. Remember that it is the RMS current times the DCR, not the average, so losses will be higher than if you use average current.
Good luck!
