Speed of monostable circuit
[Edited to include link to ONSEMI datasheet on the 2N2222 transistor, and the two plots showing how to test turn-off delay, and what that delay is.]

IF your +5volts is well filtered, and has a large bypass capacitor (100uF) to ensure the voltage does not collapse during transistor switching, then we can make some predictions.

We need to know the stable initial condition; for that, remove the timing capacitor and the 3 diodes. Given this, R2 provides base current into Q2; the ratio of R2 to R4 is such as to cause light saturation, thus Q2 collector will be about 0.7 or 0.6 or 0.5 or 0.4 or 0.3 volts, but not at zero volts.

Given Q2 collector voltage is near ground, Q1 will be off, and Q1 collector will try to rise to +5 volts, except those 3 diodes prevent that. The voltage across R1 is 5v - 4 diode drops at 2ma(estimated), or 5v - 2.5v or 2.5v. Thus 2.5mA flows into the base of Q2.

Q2 will be in hard saturation, with 5mA collector and 2.5mA into base.

Thus the DELAY comes from the need for the saturation-mode stored-charge to leave the collector-base region of Q2.

==== >>>> ADD a 1 K OHM resistor IN series with a diode, between your + input pulse and the base of Q1.

The turn-off speed of Q2 will determine most of the delay. Unless these are extremely slow transistors.

Q2 is in hard saturation, at 5mA collector current. Your turnoff base voltage is a well behaved -2 volts (set by those 3 diodes; I like this circuit).

Thus Q2 base is driven well negative, but not so far negative as to risk damaging the emitter-base junction breakdown voltage.

You need to pick a transistor, a transistor with a datasheet that makes promises about the Tdelay (the T_turn_off), and read the plots.

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https://www.onsemi.com/pub/Collateral/P2N2222A-D.PDF

Examine Figure #2 for the circuit, and Figure #6 for the time_delay: about 180 nanoSeconds if you pull the base negative by a diode-drop.

That is, the delay is 0.18 microseconds, or 0.00018 milliSeconds.