Calculate the magnitude of \$T(s = j\omega)\$ when it is not acting as a filter (which is as \$\omega \to \infty\$ for a high pass filter).
$$|T(j\omega)| = \left|-\frac{R_2}{R_1}\right| \times \frac{|j\omega|}{\left|j\omega + \frac{1}{R_1C}\right|} = \frac{R_2}{R_1} \times \frac{\omega}{\sqrt{\omega^2 + \left(\frac{1}{R_1C}\right)^2}}$$
$$\lim_{\omega \to \infty}|T(j\omega)| = \frac{R_2}{R_1}$$
since the \$\omega\$ term in the denominator dominates the \$1/(R_1C)\$ term and then the fraction with \$\omega\$ simplifies to 1. Alternatively, you can find this by inspection since the capacitor is a short circuit at high frequencies and the circuit becomes just an inverting op amp amplifier.
Next, set the transfer function magnitude equal to \$-3\text{ dB} \approx 1/\sqrt{2}\$ times the previously calculated magnitude.1 In this case the \$-R_2/R_1\$ terms cancel so you are left with
$$\frac{\omega}{\sqrt{\omega^2 + \left(\frac{1}{R_1C}\right)^2}} = \frac{1}{\sqrt{2}}$$
Solve this equation for \$\omega\$ to obtain the -3 dB frequency, which turns out to be $$\omega = \frac{1}{R_1C}$$
1The power gain of a circuit is given in dB as $$10 \log_{10}\left(\frac{P_\text{out}}{P_\text{in}}\right)$$ and the cutoff frequency is typically defined as the frequency at which the output is at half the power of the input: $$10 \log_{10}\left(\frac{1/2 \times P_\text{in}}{P_\text{in}}\right) = 10 \log_{10}(1/2) = -3.0103\text{ dB} \approx -3\text{ dB}$$
The transfer function is given in terms of voltage rather than power so we need to use the fact that \$P = V^2/R\$: $$10 \log_{10}\left(\frac{P_\text{out}}{P_\text{in}}\right) = 10 \log_{10}\left(\frac{V_\text{out}^2 / R}{V_\text{in}^2 / R}\right) = 10 \log_{10}\left(\frac{V_\text{out}^2}{V_\text{in}^2}\right) = 10 \log_{10}\left(\frac{V_\text{out}}{V_\text{in}}\right)^2 = 20 \log_{10}\left(\frac{V_\text{out}}{V_\text{in}}\right)$$
where the last equality follows from the fact that \$\log_b(x^p) = p \log_b(x)\$ and explains why we multiply the logarithm by 10 when working with power but 20 when working with voltage (or current).
Solving for \$V_\text{out}/V_\text{in}\$ at the cutoff frequency:
$$10 \log_{10}\left(\frac{V_\text{out}}{V_\text{in}}\right)^2 = 10 \log_{10}(1/2)$$
$$\left(\frac{V_\text{out}}{V_\text{in}}\right)^2 = 1/2$$
$$\frac{V_\text{out}}{V_\text{in}} = \frac{1}{\sqrt{2}}$$