Assuming an hfe of 100, you need at least 1/100th of the LED current into the transistor base. A reasonable rule-of-thumb is 1/50th to 1/20th (2%-5%) to make sure the transistor turns all the way on ;(saturates)
This is mainly what I was getting at. If I wanted a 625 mA current, then providing 1/20th of that value to the base I would need a total arduino output of 31.25 mA. I would guess that this is well within the limits of the arduino?
Do you know why it's bad practice? ...The current (around 25mA in your case) gets divided among the parallel LEDs, but it does not necessarily get divided evenly, depending on variations in LED characteristics. That means some LEDs may be brighter than others.
I understand this principle, however I'm not too worried because even if one LED pulls a lot of the current it needs to be at least 20 mA to fail. Seeing as the LEDs have the same specifications I don't imagine a single LED pulling more >20 mA, leaving <5 mA for the other 3. Maybe I'm wrong on this though 
CrossRoads - Thanks for that information. However I think I would prefer to stick with the original parallel configuration because, for me, preserving as much brightness as possible is more important than total power consumption for the moment - based on the issues mentioned by DVDdoug. I'll probably apply your advice to another project I'd like to do.
Thanks for the help.