My recommendation would be to power each LED individually. I'd use an LDO adjustable voltage regulator at around 2.7 V then try powering the LED directly without a resistor and adjust (likely lower) the voltage to the desired intensity. I would power it with a single $4-5 ≈3000 mAh Samsung 18650 Li-ion battery. You would need to be careful not to discharge the Li-ion much below 3V.
I have tested powering, without a resistor, some Samsung mid power (35 lm, 2.75V, 65 mA) LM301B LEDs at between 2.4 V (very visible) and 2.7 V (very bright).
Use Cree 5mm Green C503B-GAS-CB0F0792 and you can lower the Vf and current significantly and get the more luminous intensity as the cheap ones you choose. Lower the current to about 5 mA and the Vf will be 2.8 V or less. The extra cost for the higher efficacy LEDs (64,600 mcd) will more than pay for themselves in cost of batteries. You may even be able to go under 1 mA.
This Cree LED should draw less than 10 mA @ 2.7 V. This would still give you over 25,000 mcd.
RE: Your schematic:
If you want the eyes to be at the same 20 mA intensity, you should use a 300 Ω resistor. Better would be to combine the antenna and eyes in to a string of three LED or better yet use strings of 4 LEDs.
You could try six batteries and three LEDs in series with no resistor.
Understand that voltage of alkaline batteries will drop rather quickly. The typical alkaline AA drops from 1.5 V to 0.8V (20 mA drops to 10 mA) over its lifespan.
The correct way to power LEDs with alkaline batteries is to use a step-up/step down switching voltage regulator.
If you look at an alkaline AA datasheet you will see how quickly the voltage drops which mean the intensity will drop that fast. This is another reason to use high efficacy LEDs and a voltage controller.
You will not get the 2000 mAh because that is if the batteries are discharged to 0.8 V which would be insufficient to power three cheap LEDs in series. The Cree LEDs would still be very visible but with much less intensity than with fresh batteries.
Whereas if you use Lithium AA you will get more capacity (3500 mAh) and the voltage discharge is much flatter than alkaline so the luminous intensity will hold until near the end of its lifespan.
Your best bet would be to use two Li-ion batteries with a buck step down voltage regulator or a 3.3 V linear regulator and power each LED with its own resistor.
If you dropped the LDO voltage to about 2.7 V (for a 3 V LED) you could eliminate the resistors and use a single Li-ion cell.
A simple adjustable LDO regulator LM317
You also could use a fixed output voltage LDO in the range of 2.8-3.0 V with a resistor (≈150Ω-200Ω) on each LED for a 2 mA current giving you about a very bright 5,000 mcd. The circuit for a fixed LDO would look the same without the 2 resistors.