Increasing L may or may not increase efficiency; increase it too far likely decreases efficiency, especially if you are space-constrained.
If you keep the package size the same, larger L means more loops of thinner wire inside the inductor, hence larger DC resistance and lower current rating.
Excessively small L means higher magnetic flux ripple and possibly higher core loss, though.
I'd use the value of L suggested by the manufacturer, and only increase it if low output ripple voltage is of utmost importance (then, increase C as well, of course). Or, if you run at lower loads.
If you have the physical size and the budget to use a larger value L, and found one that is suitable per current rating, don't pick that; pick the lower value (the original suggested L) from the same series, and you'll get much lower DC resistance and extra thermal budget!
Note though that at such high f_sw, inductor AC losses (core losses + winding AC losses) are absolutely paramount. Sadly, they are not in the datasheets; some manufacturers provide web-based tools to calculate them. Often, you just pick a small ferrite-cored (shielded type!) inductor for prototyping and hope you are lucky...
What comes to your wondering of chip losses, switching losses are difficult to calculate in such integrated-FET controller, but let's calculate conduction losses for a simple example case:
Vin=12V
Vout=5V
Duty = 5/12 = 42%
MOSFET on-time = Duty
Diode on-time = (1-Duty)
MOSFET I_rms during on-time = approx. I_L_avg = 1.2A
MOSFET Rds(on) (from the datasheet) = 0.35 ohm typical
P_MOSFET_cond = Duty * I_rms_during_on^2 * 0.35 ohm = 0.42*(1.2A)^2*0.35ohm = 212 mW
Now just assume conduction losses and switching losses are 2:1, so add 50%, and we are somewhere around 300mW, well below the absolute maximum dissipation 568mW (as it should be, because this is absolute maximum rating, and at Ta=25 degC.)
From RthJ-A = 220 degC/W (note: 4-layer board), the temperature rise is around 66, say 70 degC. At T_amb = 60 degC, you are at Tj=130 degC, which is, IMHO, a barely acceptable margin. It's still 20 degC away from the typical thermal shutdown value. You could run into problems using it in a hot car on a sunny day in direct sunlight, at full 1.2A output all the time, though.
But this is a bit hand-wavy, because we can't know for sure what the switching losses are. You also need to consider the local ambient; you get roughly correct mental simulation by placing all the other components, like the diode and inductor, on the PCB, run power through them to heat them up the same amount they do in the actual circuit, and put it in the box like you finally do. This is the ambient the chip "sees". (Proper thermal simulation is much more complex, though).
The schottky diode will dissipate approximately (1-Duty)*Vf*I_L = 58%*0.4V*1.2A = 0.28W, heating it up by just 0.28W*55degC/W = 15.4 degC over ambient. So note, while the diode dissipates as much as the chip, it has way better metal leads to heatsink it. Tj ends up well below 100degC, which means the reverse leakage likely isn't a problem, but let's calculate it (at Tj=100degC) just to be sure:
P_leak = Duty * Vsupply* Ileak = 0.42*12V*6.5mA (from the curve) = 32.8mW. Insignificant here, but funnily enough, not that far from being a concern!