I looked at the pictures, it appears they are in parallel too.
the 3.3 ohm resistor will limit the current to about 500mA. What will happen is that each LED will still drop 3.3V but will not get 20mA, instead each LED will get about 5mA. This may be bright enough, depending on the LED.
If you want it brighter, and want to give 20mA to each LED, then you can use a single 3.3 ohm resistor in the power line to each level, so each level of 25 LEDs will get 20mA. But that may be too bright. Maybe it's ok at 5mA. You'll have to try it and see. So I suggest wiring it up on your bench a few different ways, before you cube it all together.
I think what I want is to wire the four layers in parallel and use a 1 Ohm 4W resistor with power being supplied by a 5W "wall wart" that is capable of supplying at least 2000mA. Does that sound right?
Yes, that would work, each LED will get 20ma then. Maybe it's too bright? or use one 3.3 ohm in each leg of 25 LEDs. You still need 2000mA. (2A) total.
Follow on question: I have a wall wart that supplies at 5V and up to 3500mA (it originally powered a USB hub), what's the connector on the end of the cable called? It's like a metal cylinder with a hole in the middle where a pin goes. I'd like to get socket for my LED cube.
Those are called coaxial power connectors. There are many different sizes, 2.1mm, 2.5mm (inside), 5.5mm, 9.1mm (outside), etc. You have to match the plug and jack and pin sizes.
http://en.wikipedia.org/wiki/Coaxial_power_connector