Here's another idea to add to the good suggestions proposed by others. If you're going to be using RGB LEDs, you should consider the power supply requirements of the device. The blue LEDs in the RGB LED package have a high forward voltage requirement, so you will pretty much need either a 3.6 volt lithium battery, or a 9 volt battery. The 3.6 volt battery is barely adequate for a blue LED - say you select your current limiting resistor based on this nominal voltage and the voltage decreases a few tenths of a volt as the battery discharges - the blue LEDs will rapidly lose brightness. You end up condemning your SO to carry around a 9 volt battery all night, or several 3.6 volt batteries in series, which is bulky.
The standard ATTiny is good to 10 mHz at down to 2.7 volts, and has two PWM channels and four AD converters. It will be slumming it just controlling a couple LEDs; there should be plenty of code space available to use the second PWM channel to implement a simple boost converter. The converter can take a barely adequate battery voltage and boost/regulate it to keep a blue LED's light output stable as the battery discharges. This opens up all sorts of possibilities for lightweight batteries; you could use 2 alkaline triple A's, for example, for really long life. There is another advantage to this scheme - you can dynamically modulate the LED power supply voltage depending upon which LEDs are being used. When the blue LEDs are on, pump the voltage up to what's needed to give them the required brightness. When they're off, drop the supply back to just above the battery output voltage for the red and green LEDS. This should give you significant power savings.
If you find that when using a 3 volt supply that the standard ATTiny browns out under some conditions when fed directly from the battery/batteries, you could try using the low voltage part, or you could actually set it up so that the direct battery voltage just gets it started, and then run the chip from the boosted supply.
Your problem is pretty straightforward. Your current limiting resistor is much too large. If your LEDs are in fact allowing as much current as you assert, the voltage across the resistor will be .08 x 30, or 2.4 volts. This leaves (at most) 0.9 volts across the LEDs, and that is not enough to allow them to produce much light at all.
You should resize your resistor, taking into account the forward voltage (Vf) of the LEDs, to allow maximum current with transistor fully on, and a transistor voltage drop of about 0.1 to 0.2 volts. Either that, or increase your source voltage.
Once you do that, you're still likely to have problems. With 5 LEDs in parallel, whichever one has the smallest Vf will hog current and glow more brightly than the others. In the worst case, this will cause it to get hotter, its Vf will drop, and it will hog even more current and get even brighter. At this worst-case limit, it will draw nearly 5 times as much current as you expect. If this level is too high, the LED may fail open, leaving the process to repeat in turn with the other 4, then the other 3, etc.
Finally, you need to examine the data sheet for your transistor and determine its current gain. This is the hfe which Ignacio referred to his comment. To make life more difficult, gain changes with current level, as you will see if you pay attention to the data sheet. But let's say that the gain is 100, which is a decent starting point for modern NPN signal transistors running at less than 100 mA. Keep in mind that, due to your large limit resistor, the current will never approach the 80 mA you think it will. Let's say 10 mA, just as a start. Then any base current above (10 mA / 100) will make no difference to the LED current, since the transistor is pulling as much current as it can, and the current is limited by the resistor and the LEDs. 10 mA/100 is 0.1 mA, or 3% of your nominal drive, and is entirely consistent with what you see.
In order to check this, fire up your circuit and connect the collector of your transistor to ground. Now measure the voltage across the 30 ohm resistor, and divide by 30, to give your total, maximum current. Divide this by 10 or so to get the base current you need. To understand why you divide by 10 rather than 100, start learning about transistor saturation.
Best Answer
You have the transistors as emitter followers so they are going to drop a lot of voltage. In a comment you said you have series resistors for each LED that are not shown in the schematic (necessary). Okay, they've been added but via an edit so all the comments about them being missing look odd .
I suggest using a single P-channel logic-level MOSFET. Low = ON in this case.
Somthing like an AOD417 which comes in a relatively easy to handle package.
At 3.6A with at least -4.5V Vgs it will be less than 55m\$\Omega\$ at 25°C so dissipation about 0.7W (will get warm, so a square inch or so of copper would be good). For more money, a https://www.vishay.com/docs/70297/sq40031el.pdf will be about 15x better.