How to calculate resistors for a 555 sawtooth with an opamp sink controlling it's frequency
The voltage at the op-amp output is -Vin  (your 1K resistors are a bit low, 10K would be better).
The base voltage (ignoring base current) is (5V/500 - Vin/4K)* (4K||500) =
Vb = 4.44V - Vin*0.111
The emitter voltage relative to the +5 when the transistor is conducting
is Vb + 0.7 -5V = 0.14V - Vin*0.111
So the emitter current ~= collector current is
Ic ~= (Vin-1.26V)*0.111mA when Vin &gt; 1.26V, approximately 0 for Vin &lt; 1.26V
The 555 charges and discharges to 2/3 and 1/3 Vcc. We can ignore the discharge time for lower frequencies since there is no resistor.
So the time to charge the capacitor through 1/3 of Vcc is:
tc = \$\frac{C\cdot 1.667V}{Ic}\$ = \$\frac{200\text{nF} \cdot 1.667V}{(Vin-
1.26V)*0.111mA}\$
Ignoring the discharge time, the frequency will be:
f = \$\frac{(Vin-
1.26V)*0.111mA}{200\text{nF} \cdot 1.667V}\$ = 333(Vin - 1.26V) Hz.
If the emitter voltage gets too close to (2/3)*Vcc the current source will saturate before it charges the capacitor sufficiently, but that won't occur unless your input voltage is more like 15V.

If you want to make an audio-range VCO that uses cheap easily available parts, I would suggest dumping the 555 and using the one on the LM324 datasheet which uses half the chip and an inexpensive BJT (or MOSFET).