Far away from a charged conductor, the field is like a point charge. Where's the point located?

The answer is that it doesn't matter.

The distance at which fields resemble that from a point charge is also the distance at which it does not matter where that point is located within the structure. The change of field due to switching the origin within the conductor will be comparable to the corrections to the point charge approximation, both arising from same order multipole terms.

Based on some of the back-and-forth I see, I think you're asking the wrong question. I think the question you want to ask is "Given a charge distribution $\rho(\mathbf{r})$, where should I place a point source so that the exact potential $\phi(\mathbf{r}) = \int \rho(\mathbf{r}')/|\mathbf{r}-\mathbf{r}'| dv'$ is most closely approximated by the potential from the point source?"

The answer is that you want to choose $\mathbf{r}_0$ such that

$\int (\mathbf{r}'-\mathbf{r}_0) \rho(\mathbf{r}') dv' = 0$

If the charge distribution is uniform, then the answer is at the centroid. The reason this is the right point is it makes the dipole moment of the difference between exact and approximate solutions go to zero. So the error in the potential is $\mathcal{O}(1/r^3)$, whereas with any other choice the error would include the dipole term, and therefore be $\mathcal{O}(1/r^2)$. (Properly setting the magnitude of the point charge accounts for the monopole term of $\mathcal{O}(1/r)$.)

Further clarification:

The choice of $\mathbf{r}_0$ that satisfies the dipole constraint above is

$\mathbf{r}_0 = \frac{\int \mathbf{r}' \rho(\mathbf{r}') dv'}{\int \rho(\mathbf{r}') dv'}$

and can be thought of a as a "center-of-charge" similar to a center-of-mass.

The multipole expansion of the potential $\phi(\mathbf{r})$ contains terms of increasing order in $1/r$

• Monopole terms decay with $\mathcal{O}(1/r)$. Any charge distributions with the same total charge within a local region have the same monopole moment. That's why a point charge with the same total charge works as an approximation, and it doesn't matter where it is, as long as it's close to the same region. With this approximation, the error between the exact potential and the approximation will be $\mathcal{O}(1/r^2)$. If $r$ is big enough, then like everyone else says, it works fine and it doesn't matter where $\mathbf{r}_0$ is.
• However, if we want, we can be even more accurate with a judicious choice of the location of the point charge. Dipole terms decay with $\mathcal{O}(1/r^2)$. Since the point source clearly has no dipole moment, picking the point $\mathbf{r}_0$ so that the exact potential has no dipole moment about $\mathbf{r}_0$ removes $\mathcal{O}(1/r^2)$ dependence from the error. This leaves only $\mathcal{O}(1/r^3)$ and higher error terms.

The multipole expansion is a useful approximation to low order (mono-, di, - quadrupole) if the diameter of the charge distribution $d$ is much smaller than the distance at which you observe the field or potential $r$.

That observation distance is with respect to an origin, which you conveniently place somewhere inside the charge distribution. Emphasis on somewhere, because wiggling with the origin at most a distance $d$ won't change much $$|r|^{-1} \approx |r+d|^{-1} \text{for d\ll r}$$ You may place the orgin, aka the source of the monopole field, anywhere inside$^1$ the charge distribution. It is not a point dictated by the theory. One would normally choose it in such a way to make higher moments vanish.

One can always make the dipole moment vanish by placing the origin at the center of charge.

$^1$ You may place it outside the (convex cone around) the charges, but then you will never get higher moments to vanish.