The 2N5551 transistors you have are weak. They perform poorly above 50mA.
Ordinary 5mm LEDs light brightly with 25mA and burn out over 30mA.
You want the transistor to fully turn on with a voltage loss of only 0.2V so the 12V supply is reduced to 11.8V for the LEDs.
You can use three white or blue 3V LEDs series then their total is 9V. The series current-limiting resistor has 11.8V - 9V= 2.8V across it and Ohm's Law calculates the resistor value to be 2.8V/25mA= 112 ohms which is not a standard value so use 120 ohms.
BUT audio is AC and the transistor turns on for only half of it so the average current in the LEDs will be half and they will look dimmed. Then reduce the value of the resistor to 56 ohms so that the peak current is 50mA and the average current is 25mA.
The weak transistor will not be able to drive more LEDs so another transistor and resistors must be used to drive 3 more LEDs.
The series base resistor must have a current of 5mA but we cannot calculate its value without knowing the output voltage of your amplifier.
Ordinary 5mm LEDs light brightly with 25mA and burn out over 30mA.
You want the transistor to fully turn on with a voltage loss of only 0.2V so the 12V supply is reduced to 11.8V for the LEDs.
You can use three white or blue 3V LEDs series then their total is 9V. The series current-limiting resistor has 11.8V - 9V= 2.8V across it and Ohm's Law calculates the resistor value to be 2.8V/25mA= 112 ohms which is not a standard value so use 120 ohms.
BUT audio is AC and the transistor turns on for only half of it so the average current in the LEDs will be half and they will look dimmed. Then reduce the value of the resistor to 56 ohms so that the peak current is 50mA and the average current is 25mA.
The weak transistor will not be able to drive more LEDs so another transistor and resistors must be used to drive 3 more LEDs.
The series base resistor must have a current of 5mA but we cannot calculate its value without knowing the output voltage of your amplifier.