How does the voltage go negative on the BJT base in this simple flasher?
To gain some insight, consider the following simple circuit:







simulate this circuit &ndash; Schematic created using CircuitLab

Assume the switch has been in the top position for some time such that the voltage across the capacitor is, e.g., 8V with the top terminal the more positive.  The voltage at the anode of the diode is about 0.7V.

Then, the switch is moved to the lower position.

Now the top terminal of the capacitor is connected to the 0V reference so the voltage there is 0V.

But, since there is no path for the capacitor to discharge (the diode blocks any discharge current), the voltage across the capacitor must still be 8V with the top terminal the more positive.

Thus, the anode of the diode (connected to the lower terminal of the capacitor) must now be -8V.

We've produced a negative voltage by first charging the capacitor to a particular voltage and then we switch the circuit into a different configuration where the voltage across the capacitor is used to produce a negative voltage.

In your circuit, the left most transistor acts as the switch, grounding the left-most capacitor terminal thus producing a negative voltage on the base of the right-most transistor.