3.7 720mAh lipo thru TIP120 as a relay -- no current?
The TIP120 is a Darlington transistor pair. Internally it looks like this:-







simulate this circuit &ndash; Schematic created using CircuitLab

The advantage of a darlington pair over a regular bipolar transistor is much higher current gain. Since the output current of the first transistor feeds directly into the Base of the second transistor, the total current gain is equal to the product of the two, eg. if Q1 and Q2 both have a gain of 100 the total gain is 100 x 100 = 10,000. 

However this comes at the cost of higher voltage drop. Between the 'Base' and 'Emitter' there are two Base-Emitter junctions each dropping ~0.6V, so it needs ~1.2V to turn on. Q1 works off the voltage between Q2's Collector and Base, so the 'Collector' can't go below about 1V. 

Another potential problem is that the very high current gain can make it susceptible to small leakage currents. The 8k and 120&ohm; resistors make sure these leakage currents don't cause the transistors to turn on when there is no input to the 'Base', but they also reduce gain at low current. 

Luckily these disadvantages won't affect you, because the TIP120's current gain peaks at ~1-4A which is close to the expected current draw of your glow plug (3~5A), and the battery has enough excess voltage to still get the required 1.5~2V at the glow plug. However you do need to supply enough 'Base' current to turn the Darlington transistor on fully. 

To get lowest Collector-Emitter voltage drop at 5A a Base current of ~5mA is required. At this current Base-Emitter voltage is about 1.8V. The Arduino puts out 5V so the 2.2k&ohm; resistor has 5-1.8 = ~3.2V across it. 3.2V/2.2k&ohm; = ~1.5mA, which may not be enough to turn the transistor on fully. To get 5mA Base current you need a resistance of 3.2V/5mA = 640&ohm; or less (560&ohm; or 470&ohm; would be OK).  

Note that if the glow plug is drawing 5A the transistor has to dissipate ~1.5V * 5A = 7.5W. The TO220 package has a Junction−to−Ambient thermal resistance of 62.5&deg;C/W, so it would heat up by 7.5 * 62.5 = 468&deg;C internally. This would quickly burn it out, so you should attach a large heat sink to keep the temperature down. 

The transistor + heat sink may be larger than a relay! A 'logic level' power MOSFET could do the job with much lower voltage drop and power loss. Then you might not need any heat sinking, and you could apply PWM to reduce average glow plug current so the higher voltage doesn't burn it out.