Is chromate a suitable indicator for the titration of Ag⁺ with Cl⁻?
You got the solubility part reversed.
The solubility of $\ce{AgCl}$ is lower than the solubility of $\ce{Ag2CrO4}:$

$$s(\ce{AgCl}) = \sqrt{K_\mathrm{sp}(\ce{AgCl})} = \sqrt{\pu{1.8E-10 mol2 L-2}} = \pu{1.34E-5 mol L-1}$$

$$s(\ce{Ag2CrO4}) = \sqrt[3]{\frac{K_\mathrm{sp}(\ce{Ag2CrO4})}{4}} = \sqrt[3]{\frac{\pu{1.1E-12 mol3 L-3}}{4}} = \pu{6.50E-5 mol L-1}$$

Therefore, if the $\ce{AgNO3}$ solution is gradually added to the solution containing the both $\ce{Cl-}$ and $\ce{CrO4^2-}$ ions, then initially the formation of a sparingly soluble $\ce{AgCl}$ salt occurs.
After the $\ce{Cl-}$ ions are almost completely isolated in the form of $\ce{AgCl},$ the $\ce{Ag2CrO4}$ precipitation starts to occur, signifying the equivalence point is reached.

The same reasoning can also be applied to the titration of even less soluble silver bromide $\ce{AgBr}$ with $K_\mathrm{sp}(\ce{AgBr}) = \pu{5.3E-13}$ (try it yourself).

Note, however, that while using Mohr's method it's imperative to titrate halide salts solutions with $\ce{AgNO3}$ and not vice versa.
Otherwise the precipitation condition 

$$c(\ce{Ag+}) · c(\ce{Cl-}) &gt; K_\mathrm{sp}(\ce{AgCl})$$

will be overridden by

$$c(\ce{Ag+})^2 · c(\ce{CrO4^2-}) &gt; K_\mathrm{sp}(\ce{Ag2CrO4})$$

due to high concentration of silver ions in solution, favoring silver chromate precipitation (note squared term $c(\ce{Ag+})^2$) and thus shifting the equivalence point.