Questions on this high frequency sawtooth generator
the inverter goes low, thereby letting the C_T discharge through the D_2 diode. But after that where does it drain?


When the inverter output is low, the NMOS transistor of the inverter's output stage effectively connects the output to ground. You can look in the datasheet for the maximum sinking current to get an estimate of how fast it can discharge \$C_T\$.


  What can I do to make the discharge faster so I can make the negative edge more vertical?


Reduce \$C_T\$.

You'd then have to increase \$R_T\$ to compensate if you want to keep the rising edge time the same.

To go really fast, the diode might become important and you might want to look for a faster-switching diode (but I haven't looked at the 1N4148 specs---it might already be fast enough).

Edit: I checked the 1N4148 specs. Fairchild specs it with 4 ns reverse recovery time. This spec doesn't slow down the maximum switching rate of your circuit, but it does limit how fast you can go and still maintain the sawtooth waveform shape (slow rising edge and fast falling edge). With this diode if you try to make your sawtooth faster than maybe 10 MHz, you'll probably see noticeable distortion due the diode's reverse recovery feature. The initial ~4 ns of the rising edge will be faster than the desired ramp rate.


  Can I go for hundreds of MHz?


I would guess that 100 MHz is doable. Much higher than that is probably quite challenging if it's possible at all. 

With 2.2 ns propagatin delay, 227 MHz will be the absolute limit.


  Is the high charging current (for R_T being so low) too much for the C_T to charge at a consistent time?


I'm not sure what you mean by this question. You will probably see the discharge time vary a bit if the IC's temperature changes. The charging time probably has more to do with the stability of the Schmitt trigger's thresholds, the stability of Vcc, and the stability of \$R_T\$.