How to simplify these system of equilibrium equations
Since the equilibrium constant for the formation of the complex ion is very large, I assume that $\ce{[Ni(CN)4^{2-}] &gt;&gt; [Ni^{2+}]}$


From the comments:


  [comments:] That assumption is incorrect. The equilibrium constant seems large, but the exponents are high, so it is misleading. In fact, more of the cyanide is in the form of HCN than in complex with nickel, and there is very little free cyanide. 


OP replied:


  How may I make a better assumption to solve this problem?


In the absence of nickel, the pH is around 5 and HCN is the major species, with the concentration of cyanide in the micromolar range (standard weak acid problem). When you add nickel, the concentration of cyanide gets lowered further. Because the free cyanide concentration contributes little to the total, we can combine the two reactions to see whether there is more complex or more HCN and to determine the pH:

$$\ce{4HCN + Ni^2+ &lt;=&gt; Ni(CN)4^2- + 4H+}$$

The equilibrium constant is 

$$K_\mathrm{combined} = \pu{5.8e−16}$$

So for this reaction, reactants are favored. HCN and nickel concentrations hardly decrease compared to the initial concentrations. If we call the concentration of the complex $x$, we get:

$$K_\mathrm{combined} = \frac{x (4x)^4}{[\ce{HCN}]^4 \cdot [\ce{Ni^2+}]}$$

Solving for $x$, we get $\pu{2.24e−5}$, confirming that reactants are favored. From this we can calculate the pH as

$$\mathrm{pH} = -\log 4x = 4.05 $$

Checking the solution

Upon adding nickel, the pH turned a bit more acidic. That makes sense as a micromolar concentration of nickel complex formed, releasing some more protons from HCN. We can calculate the cyanide concentration from either one of the given equilibrium constants, and verify that it is miniscule. Then, we can verify that all reactions are at equilibrium (at pH 4, hydroxide is a minor species as well, and has little impact on the exact solution).