Question about relation between two time constants
I think I understood your question now. I will reformulate it in more 'chemical' terms and give you my solution. Not nicely formatted for now, I will do that later if I find the time.

You have a cell with $G_{max}$ gates on it. They can open and close reversibly, depending on a voltage V. Closed gates open with rate constant $\alpha(V)$, and open gates close with rate constant $\beta(V)$.

This implies one differential equation. If $G_o$ is the number of open gates at time t, and $G_c$ is the number of closed gates at time t, then:

$\frac {dG_o}{dt} = \alpha G_c - \beta G_o$

And as the total number of gates is $G_{max} = G_o + G_c$:

$\frac {dG_o}{dt} = \alpha (G_{max} - G_o) - \beta G_o$

At steady-state, i.e. when $G_o$ stops changing, you have:

$\frac {dG_o}{dt} = \alpha (G_{max} - G_o) - \beta G_o = 0$

which, solved for $G_o/G_{max}$, yields:

$\frac {G_o}{G_{max}} = \frac {\alpha}{\alpha + \beta}$

The ratio $G_o/G_{max}$ is the probability that a single gate is open. As you say, a 'channel' is open when 4 of its gates are open. If we consider each gate being open or closed independently, the probability that 4 gates in a channel are open is indeed $(G_o/G_{max})^4$

What you measure is the time it takes, when the voltage is abruptly changed from an initial value to a final value, for the number of open channels to go from an initial value to 'equilibrium'.
This time is in theory infinite, but in practice it is probably the time it takes for the number of open gates to go from its steady-state value at voltage $V_1$, to a value sufficiently close to the new steady-state value at voltage $V_2$.

If so, it is sufficient to solve the above differential equation for $G_o$ with initial conditions $(t=0,G_o=G_{o,V_1})$. The solution is:

$\frac {G_o(t)}{G_{max}} = \frac {G_{o,V_1}}{G_{max}} \cdot e^{-{(\alpha+\beta)t}} + \frac {G_{o,V_2}}{G_{max}} \cdot [1-e^{-{(\alpha+\beta)t}}]$

where:

$\frac {G_{o,V_i}}{G_{max}} = \frac {\alpha(V_i)}{\alpha(V_i) + \beta(V_i)}$

The difficulty I see is the arbitrary nature of the 'time' you consider sufficient to reach steady-state. I would rather fit time points, i.e. number of open gates vs time. And my earlier suggestion to use numerical fitting may still be valid, actually.