The 2N2222 might be a better choice - inexpensive, commonly available, handles the current, and overall a good choice for switching.
The spec you want to look at most Icmax, or sometimes just Ic (The 'C' being a subscript) which is the maximum current you'd normally be able to put through a fully turned on (saturated) transistor.
The 2N2222 apparently is popular enough to get its own web domain http://2n2222datasheet.com/ where I found several PDF spec sheets. I see (pun not intended) that Ic is 600mA - you could use one transistor to drive all your LEDs.
Another spec to pay attention to is beta - the current gain. If you're switching 400mA and the transistor has a beta of, say , 100 then you'll need to supply 400mA/100 = 4mA to the base from your digital output. Beta isn't very consistent from transistor to transistor, even of the same type. Just make sure the math works out for the lower end of the beta range when choosing a resistor for the base.
Practically all the other specs aren't of as much importance, not a your low 5V supply, unless you're going to drive the LEDs very fast, e.g. a few MHz.
You have a minimum current gain given as 1000 in the datasheet, so this means 1mA into the base should result in at least 1A from collector to emitter, assuming the supply can provide this.
The calculation for current into the base (assuming the Arduino pin outputs 5V high) and we take the maximum base-emitter voltage from the datasheet is:
(Vpin - Vbe) / Rbase = (5V - 2.5V) / 1kΩ = 2.5mA;
It looks like your multimeter has quite a high burden voltage (too high a high resistance used for current shunt, so you get a voltage drop across it which affects things) hence the readings being out on the base current and the difference between the supply-collector reading and emitter-ground reading (which should be practically the same - the emitter-ground reading only has the base current added to it, which is tiny compared to Ic)
The darlington transistor has a high saturation voltage (higher than a normal transistor), so a higher voltage supply is preferable for reasonable results, and gain also drops at saturation. In any case, controlling the current in this way is not very practical, since the gain can vary widely between parts, with temperature, etc. Try adding another battery or two, adding a small value collector resistor (capable of the wattage it needs to handle to control max current accurately), and lowering the base resistor to around 200Ω.
If you want to learn about the base current vs gain and saturation relationship, try using a higher supply voltage you know is capable of the current you are testing with, adding a potentiometer (wired as a rheostat) at the base, a small value resistor to give you a known maximum current, and plotting the base current vs collector voltage/current. If you do the calculation you should be able to see how the gain varies and drops approaching saturation. You should get plots similar to the datasheets.
Doing the above in SPICE is also another option if you don't have enough test equipment to make things easy.
Best Answer
The TIP120 does not need 120mA at the base for normal operation, that's the absolute maximum rating, above which you don't want to go.
The spec you are mostly interested in is the hFE (current gain), which for a darlington is very high, since it's two transistors connected in a way so the current gains multiply. For the TIP120 it's given as minimum 1,000 (compare with a typical 200 for a single bipolar transistor)
Also important are the max collector current (5A) and the collector emitter voltage (60V)
The main disadvantages are that the base-emitter voltage is doubled compared to a single transistor (~1.4V), and the saturation voltage is higher (typically ~0.8V compared to ~0.2V at low currents)
These points are rarely a problem for a simple switch driven from a micro pin. At higher collector-emitter currents though, the Vsat rises and can interfere with desired operation and cause problems with dissipation.
For example, in the TIP120 datasheet note that at 3A Ice, Vsat is given as 2V, but at 5A it has risen to 4V. That's 20W of dissipation, a lot to heat to try and get rid of to keep the temperature down. So when switching a large current you need to take these factors into account, and maybe decide to look at a more suitable part (e.g. logic level, low Rsdon power MOSFET)
Since we have a gain of 1000, we hardly have to draw anything from the micro pin. Let's say we want to switch 1 Amp:
1A / 1000 = 1mA into the base needed.
If we have a drive voltage of 5V, then we subtract the Vbe from the drive voltage and divide by the current:
(5V - 1.4V) / 1mA = 3.6k resistor. To give it a bit of leeway select something a bit smaller like 2.2k. This still only draws ~1.6mA.
I wouldn't read too much into the different prices - the price of components is often dictated by how popular they are, the more they sell the less they cost. If you see better specs at a cheaper price, go for it ;-)
You can some pretty odd prices when the component is scarce/new/obsolete - I saw a 10uF ceramic capacitor priced at £7.50 (qty 1) on Farnell the other week...