Expanding upon my comment above, if you only need a few digits of accuracy you can probably use the method of stationary phase. We can follow the procedure on Wikipedia. We can write the transform as follows: $$ f(t) = \int_{-\infty}^{\infty} g(\omega)e^{i\phi(\omega,t)}d\omega, $$ where $\phi(\omega,t)=C\omega^2-\omega t$. The principal contribution of the integrand to the value of the integral comes in the vicinity of the stationary point of the phase function $\phi$, which can be found by solving $$ \frac{\partial\phi}{\partial\omega}=0 \rightarrow \omega_0=t/2C. $$ The Taylor series expansion of $\phi$ in the vicinity of the stationary point $\omega_0$ is given, to quadratic degree, by $$ \phi(\omega,t) \approx \frac{t^2}{4C} + C (\omega-\omega_0)^2. $$ Substituting this into the transform integral, we find $$ f(t) \approx \left|g\left(\frac{t}{2C}\right)\right| e^{it^2/4C} \int_{-\infty}^{\infty} e^{iC(\omega-\omega_0)^2}d\omega. $$ Substituting an approximation with an error term that goes as $\mathcal{O}(1/C)$ for the remaining highly-oscillatory integral, we find that $$ f(t) \approx \left|g\left(\frac{t}{2C}\right)\right| e^{i(t^2/4C\pm\pi/4)} \sqrt{\frac{\pi}{C}}. $$

Using $g(\omega)=e^{-\omega^2}$ and $C=10^6$, this approximation is accurate to 3 digits up to $t=100$ (verified with Wolfram Alpha).

Expanding upon my comment above, if you only need a few digits of accuracy you can probably use the method of stationary phase. We can follow the procedure on Wikipedia. We can write the transform as follows: $$ f(t) = \int_{-\infty}^{\infty} g(\omega)e^{i\phi(\omega,t)}d\omega, $$ where $\phi(\omega,t)=C\omega^2-\omega t$. The principal contribution of the integrand to the value of the integral comes in the vicinity of the stationary point of the phase function $\phi$, which can be found by solving $$ \frac{\partial\phi}{\partial\omega}=0 \rightarrow \omega_0=t/2C. $$ The Taylor series expansion of $\phi$ in the vicinity of the stationary point $\omega_0$ is given, to quadratic degree, by $$ \phi(\omega,t) \approx \frac{t^2}{4C} + C (\omega-\omega_0)^2. $$ Substituting this into the transform integral, we find $$ f(t) \approx \left|g\left(\frac{t}{2C}\right)\right| e^{it^2/4C} \int_{-\infty}^{\infty} e^{iC(\omega-\omega_0)^2}d\omega. $$ Substituting an approximation with an error term that goes as $\mathcal{O}(C^{-1})$ for the remaining highly-oscillatory integral, we find that $$ f(t) \approx \left|g\left(\frac{t}{2C}\right)\right| e^{i(t^2/4C+\pi/4)} \sqrt{\frac{\pi}{C}}. $$

Using $g(\omega)=e^{-\omega^2}$ and $C=10^{12}$, this approximation is accurate to 6 digits up to $t=1000$ (verified with Wolfram Alpha).