Does it matter at all if the LEDs get less than 15mA?
The 2N2222 has a saturation voltage of 0.3 V at 150mA. It may be less at 15mA.
That means there will always be a voltage drop across the transistor.
Taking the typical forward voltage of 1.7V for a red LED, a 5V supply, and a 75 ohm resistor, I get a current of 44 mA. (\$I_{C} = \frac {5V-1.7V} {75 ohms} = 44mA \$ - that's 14.67mA per LED.)
Now assuming 0.3V across the collector-emitter junction, and using the rest of the numbers from my example, I get 40 mA. (\$I_{C} = \frac {5V-1.7V-0.3V} {75 ohms} = 40mA \$ - that's 13.3mA per LED.)
That's a difference of just over 1mA for each LED.
Does that difference matter? No, not for a typical LED indicator circuit.
It especially doesn't matter for your circuit.
You would normally use one resistor for each LED. Using one resistor will potentially allow one LED to "snarf up all the juice" - one LED may light and the others not, or they could all receive different amounts of current.
LEDs aren't identical (though they can be very close to identical if they come from the same production batch.) Small differences can cause the LEDs to behave just a bit differently. If you are lucky, your LEDs will be a little differently bright. If you are unlucky, one will light up (and maybe burn out) and the others will be dark.
It therefore really doesn't matter much if you get a milliampere more or less - the LEDs won't be divying it up equally, anyway.
You should do something like this:

simulate this circuit – Schematic created using CircuitLab
Each LED has its own resistor, and each gets about \$I_{C} = \frac {5V-1.7V-0.3V} {225 ohms} = 13.3mA \$ - and each resistor will get its 13mA, pretty much regardless of what the other LEDs do.