If I add some KOH, where will the reaction shifts?
The equilibrium will shift to the left, as $\ce{Ni^2+}$ is removed from the equilibrium by the precipitation of $\ce{Ni(OH)2}$:

$$\ce{Ni^2+ + 2 OH- \rightleftharpoons Ni(OH)2\downarrow}$$

The precipitate forms when $[\ce{Ni^2+}]$ and $[\ce{OH-}]$ are sufficiently high to exceed the solubility product of $\ce{Ni(OH)2}$ ($K_{sp}=5.48 \times 10^{-16}$ at 25 °C, source). The concentration of $\ce{Ni^2+}$ before addition of $\ce{KOH}$ depends on the stability constants of the ammin complexes (there are several species in equilibrium with each other, as the ligands are exchanged stepwise, and not all at once). Nickel hydroxide will dissolve again when $\ce{NH3}$ is added, as this shifts the equilibrium of ammin complex formation to the right.