The brightness of an LED is roughly proportional to the current, though it falls short of this linear relationship as the current approaches the maximum. Or, to put it another way, LEDs are slightly more efficient when run at maybe 1/2 the max current, where they will put out more than 1/2 the light that they would at full power. Of course that is a tradeoff with the cost of the LEDs.
The simplest way to get more battery life for your project is to run 2 LEDs in series. Here is the difference. If you run 2 LEDs in parallel, they will draw twice the current of 1 LED. If you run them in series they will draw the same current as 1 LED while producing twice as much light.
Your current situation is as follows:
3 LEDs run in parallel
3.7V battery
2.7Ω resistor for each LED
Since this will not run them at full current, the voltage will be somewhat less than the 1.4V they drop at 1.4A, let's estimate this at 1,2V
So:
V = I * R
(3.7 - 1.2) = I * 2.7
(3.7 - 1.2) / 2.7 = I = 0.92 A
So you the power you are putting into the LED is:
1.2 * 0.92 * 3 = 3.3W
The power you are drawing from the battery is:
3.7 * 0.92 * 3 = 10.2W
So you are getting approximately 33% efficiency with your circuit.
Let's see what would happen to get the same light with 2 LEDs in series:
The current would have to be 3/2 of the current you are supplying to the 3 LEDs, so:
I = 3/2 * 0.92 = 1.38A
How clever, that is almost exactly the 1.4A that your LEDs are rated at!
So now we have:
V = 1.4 V
I = 1.38 A
We can calculate the resistor as:
V = I * R
(3.7 - 1.4) = 1.38 * R
R = (3.7 - 2.8) / 1.38 = 0.65Ω
So, now, the power you are putting into the LEDs is:
2.8 * 1.38 = 3.9 W
And the power drawn from the battery is:
3.7 * 1.38 = 5.1 W
So you are getting about twice the light for 1/2 the power or about 76% efficiency, which is really not that bad.
I would actually halve the current and use two chains of 2 LEDs in series / parallel, which would be a little more efficient in light output because LEDs are more efficient at lower currents. Plus it spreads out the heat making it easier to heat sink them. (You did know that you have to have a heat sink for these LEDs, right?)
Compare to a switch mode converter that can be up to 90% efficient, but is much more complex.
Bob