12 volts touching any pin will definitely kill a Teensy 4.0.
Normally the ground wires should be directly connected. Best practice is to use a 5V buffer chip between Teensy and the LED input (12V LEDs usually expect a 5V signal).
A resistor between the buffer chip and LEDs can help protect it, but accidental shorting 12V is a difficult failure to protect against hardware damage. A higher value resistor limits the current, but also gives a weak drive of the signal onto your cable. Usually 100 to 220 ohms is best to get a good signal.
When the fault happens, current will flow back through the resistor. It has to go somewhere. Most chips have tiny diodes between the pins and power supply. If the 5V buffer chip you use is made that way, the current flows back into the 5V power. If the current is less than whatever total power consumption the board needs, it just feeds the 5V power. But if it is more current that the circuitry needs, it will raise the 5V power line higher, possibly destroying everything connected to 5V power.
One common way to protect circuits is with a zener diode. For example, if you use a 5.1V or 5.6V zener between the buffer output and GND, and then a resistor between that signal and the output, when the signal gets accidentally connected to 12V, the zener diode gives a path for the current to flow to GND (and heat up the zener diode). If you have a 100 ohm resistor and 5.1V zener, the current could be nearly 70 mA. If the diode becomes 6V (zeners are far from ideal) under that condition, you'll have 60 mA flowing. That's roughly 0.36 watts dissipated in the resistor and 0.36 watts in the zener diode, so at the very least you should use parts rated for 1/2 watt to avoid them overheating or burning during the fault condition. The zener won't strictly limit the voltage to 5.1V, so you might add another 100 ohm resistor between the buffer and zener. Then you have a total of 200 ohms between the buffer and wire to the LEDs. If the zener allows 6V, then the other resistor limits the current flowing back into the 5V power through the diode inside the chip to approx 4mA (assuming 0.6V forward voltage on that diode inside the chip before it begins to really conduct).
When building this, remember a lot of power dissipates in the final resistor and zener diode during the 12V fault condition. Those parts will get hot if the fault condition is sustained for more than a second or two. Even if they're both rated for 1 watt, they still can get very hot. Power rating on parts tends to be pretty optimistic about natural convention cooling (remember all datasheets are written by marketing people with the goal of convincing you to buy the part), so keep in mind the nature of needing to get the heat removed if you're packing this circuit tightly into an air tight plastic box.
