As someone else has said, this would be pretty easy to do with a microcontroller, so I thought I'd try a discrete logic approach for fun.
I'm using a CD4022 octal counter as a state machine.
I'm also including a clock circuit, which provides a free-running 1/4 second clock pulse (or whatever you decide the timing should be) going into the CLK input of the counter. It uses a 555 timer configured in an astable mode. Here's a calculator which I used to figure out the required component values.
Here's how it works. When the brake is off, BRAKE will be at ground and the N-channel MOSFET Q3 will be off. So the pullup resistor R4 will keep the counter reset line RES high. Q0 will be high, which is not connected to anything. The other two MOSFETs will be off, so the BRAKE_LED lead will not have +12v on it.
When the the brake is on, and BRAKE line is at +12v, Q3 will be turned on, bringing the RES line low and allow the clock input to increment the counter. The Q7 output of the counter is also low, which is fed into the CLR line (called !CLOCK_ENABLE on some datasheets for the 4022). Sorry about the confusion between the Q outputs on the counter and the MOSFET designations; can't really do anything about that.
The counter counts to 1, and Q1 output goes high. This turns on the N-channel MOSFET Q2, which then turns on the P-channel MOSFET Q1, which causes +12v to be applied to the BRAKE_LED line. after 1/4 of a second, the counter advances, Q1 output goes low and the Q2 output goes high. Now MOSFET Q1 is no longer on, so the BRAKE_LED line goes back off.
This repeats for two more cycles (Q3/Q4 and Q5/Q6), generating three on/off pulses altogether. Finally the counter advances to Q7, which stops the counter from advancing further because the CLR/!CLOCK_ENABLE line is now asserted. Meanwhile Q7 high also keeps the Q2 MOSFET on through the diode, which is there so the earlier assertions of the Q2 gate lead don't stop the counter by feeding back into Q7.
When the brake is released, BRAKE goes back to ground, turning off the Q3 MOSFET and resetting the counter. All FETs are once again off, and the BRAKE_LED line is inactive.
The BRAKE lead doesn't need to be debounced, since if it bounces back from +12v to ground a few times 10's of milliseconds apart, it will just keep resetting the counter and starting over which will not be visible since the counter only increments every 1/4 second.
The parts are all available in through-hole packages from Digi-Key. VN10LP is an N-channel MOSFET in a TO-92 package. ZVP2106A is a P-channel FET also in a TO-92 package. The two ICs, LM555N and CD4022, are in DIP packages and can both be operated directly off of 12v so there is no need for a step-down regulator. All caps should be rated at 25v or higher. Resistors can be either 1/4 or 1/8 watt; none of them carry any appreciable current.
Best Answer
LED Strip Basics
As you might be aware, these LED strips come as parallel groups or 3 series LEDs with one series resistor. Connecting 12V to the main connectors is all it takes to light them up. They can be cut apart, but only in groups of three at the appropriate markings on the strips. The embedded resistor value varies for different types of strips (LED color, manufacturer, etc).
simulate this circuit – Schematic created using CircuitLab
Replacing the bulbs in your car with DIY LEDs strips may not be legal. Car indication lights have to be within a specific brightness range. You to need to ensure the LEDs have the correct candella rating to be used as tail or indicator lights.
Practical Considerations
Dimming an entire bank of LEDs by placing a resistor in series with the power line is not a good idea. A lot of power will be dissipated in this resistor, as in the dropped voltage multiplied by the entire LED bank current.
Some car turn signals operate a bit strange... the ON resistance of the indicator bulb actually determines the speed of the blinker. This is part of what causes a blinker to double in speed when one of the bulbs is out. Replacing the bulb with LEDs may change the speed of the blinker if the resistance is not matched. There is also the factor of heat. LEDs don't produce very much heat, meaning your light housings could frost over in cold weather - something prevented by the heat from standard bulbs.
Also, powering the strip with 7V will probably not produce any light at all. Bright white LEDs typically drop about 3V apiece just to barely turn on. That means you need at least 9V for the LEDs plus a bit more for the embedded resistors. The extra source voltage is dropped by the embedded resistors, and this is also what determines the LED current: I_LED = [V_source - (3 * V_LED)] / R. LED brightness is determined by forward current; however, the forward voltage also changes with the forward current. A curve relating the two should be available in the LED datasheet.
How to Do It (Your Way)
If you really want to move forward with this idea, a standard rectifying diode is a good bet, but the actual part is determined by how much current will be used by the LEDs - the diode will need to be rated for at least that of the entire LED array. Since the signal won't be switching quickly ( turn signals are usually 1 - 2 Hz) that is not a factor.
Finding the necessary series resistance is a bit trickier, but doable. You will need to know how much current to pass through the LEDs to get the dimmer output you desire, then add up the LED voltages at that forward current plus the dropped voltage across the embedded resistor (V = IR). How ever much voltage is left will need to be dropped by the additional series resistor. However, keep in mind, that this resistor will have the entire LED bank current going through it...