How to understand the relationship between the filter time constant and the half-power cutoff frequency
If you have a first order high-pass filter with transfer function

$$H(s)=\frac{s}{s+\omega_0}\tag{1}$$

then the squared magnitude of its frequency response is given by

$$|H(j\omega)|^2=\frac{\omega^2}{\omega^2+\omega_0^2}\tag{2}$$

Consequently, for $\omega=\omega_0$ we have

$$|H(j\omega_0)|^2=\frac12\tag{3}$$

and $\omega_0$ clearly is the $3$-dB (half-power) cut-off frequency.

In the time domain, we could look at the step response of that system. The step response corresponding to the transfer function $(1)$ is

$$s(t)=e^{-t/T}u(t),\quad T=\frac{1}{\omega_0}\tag{4}$$

where $u(t)$ is the unit step function. So the time constant $T$ and the $3$-dB cut-off frequency are inverses of each other. This is what you get from the math (Fourier transform).

I'll try to give you some intuition and insight into this result. In general a high-pass does not pass DC (i.e., a constant value), so after the step has been applied at the input, the output must decay to zero because the value of the step function  is constant after the step has occurred. The higher the cut-off frequency $\omega_0$, the faster the output must decay to zero, because the filter suppresses low frequencies better than a filter with a lower cut-off frequency. So a high value for $\omega_0$ implies a smaller time constant $T$. You can also look at it from the other side. Assume an extremely low cut-off frequency $\omega_0\approx 0$. This means that the filter hardly suppresses anything, i.e., it leaves the input almost unchanged. So at the output you will observe almost the same as at the input, i.e., a step. Only after a very long time will the output decay to zero (because DC is suppressed, after all). So a small value for $\omega_0$ corresponds to a large value of $T$.