To expand a bit on @kingledion 's answer:
The angular resolution of the naked eye is about $\frac{\pi}{10800}$, or one arc minute.
You mentioned a civilisation on the level of ancient Rome, which had walls of around $10\left(m\right)$ high.
From basic trigonometry we know that $\tan{\frac{\pi}{10800}}=\frac{10\left(m\right)}{d\left(m\right)}$, so $d\left(m\right)=\frac{10\left(m\right)}{\tan{\frac{\pi}{10800}}}=34377.5\left(m\right)$, or more generally:
$$d\left(m\right)=\frac{h\left(m\right)}{\tan{\frac{\pi}{10800}}}=3437.75 h\left(m\right)$$
This means that on a clear day, the Maximum distance at which the average human eye can resolve an object of a $5\left(m\right)$ radius is around $34\left(km\right)$.
However, at this range, the walls would probably just look like a thin line of different colour from the rock.
The formula scales linearly, so if the walls could be made out properly with 4 "pixels", the distance would be about $8.6\left(km\right)$
This isn't all though, after all, you mention that the city is on a $~2000\left(m\right)$ high plateau, this changes the formula.
The angle at which we are looking is now $tan^{-1}{\left(s+\frac{h}{2}\right)}=\alpha=\tan^{-1}{\frac{2005\left(m\right)}{d\left(m\right)}}$
We need to multiply the distance with the cosine of this value. So:
$$d\left(m\right)=\cos{\left[\tan^{-1}{\frac{\left(s+\frac{h}{2}\right)\left(m\right)}{d\left(m\right)}}\right]}\frac{h\left(m\right)}{\tan{\frac{\pi}{10800}}}$$
Solving this numerically with Mathematica:
Solve[Cos[ArcTan[(s + h/2)/d]] h/Tan[Pi/10800] == d]
d -> 1/2 Sqrt[-h^2 - 4 h s - 4 s^2 + 4 h^2 Cot[\[Pi]/10800]^2]
and
With[{s = 2000, h = 10},
Solve[Cos[ArcTan[(s + h/2)/d]] h/Tan[Pi/10800] == d]] // N
d -> 34318.9
Plot[{d, Cos[ArcTan[2005/d]] 10/Tan[Pi/10800]}, {d, 0, 35000}]

The distance for a variable height of the plateau (s) is givien by:
$$\sqrt{-25-10s-s^2+100\cot{\left(\frac{\pi}{10800}\right)}^2}$$
