Nodal Analysis #2 - Where is my mistake?
You are making this far too difficult. 

There's no need to insert all the currents in the diagram and write expressions for each. Better is, at each node, write KCL directly using: sum of currents away = 0.

Thus, node \$\small V_A\$:
$$\small 4+(V_A-V_B)+\frac{V_A}{j}=0$$

node \$\small V_B\$:
$$\small (V_B-V_A)+\frac{(V_B-V_C)}{-j}+2(V_C-V_B)=0$$

node \$\small V_C\$:
$$\small -4+\frac{V_C-V_B}{-j}+\frac{V_C}{2}=0$$
Solve these simultaneously for \$\small V_C\$, then divide by \$\small 2\$ for \$\small V_O\$:

$$\small V_O=\frac{V_C}{2}=\frac{4}{5}(3-j)=2.53\large \angle \small-18.4^o$$

I see that your model answer gives a phase angle of \$\small 71.6^o\$. I'll check my analysis again as your answer is arctan(3), whereas mine is arctan(-1/3).