How many pi bonds are in ["Ag"("CN")_2]^(-)?

A $\pi$ bond is always made by sidelong interaction between two orbitals. If it is between two $p$ orbitals, then two out of two lobes overlap. If it is between two $d$ orbitals, then two out of four lobes interact.
You can imagine it as the second and third bonds in a triple bond. ${\text{CN}}^{-}$ has a $- 1$ charge, as is implied by the $+ 1$ charge of silver and the overall $- 1$ charge of dicyanosilver(I).
The structure of ["Ag"("CN")_2]^(-) looks like this:
$: \text{N"-=stackrel((-))("C"):-stackrel((+))("Ag")- :stackrel((-))("C")-="N} :$
So this complex will have four $\setminus m a t h b f \left(\pi\right)$ bonds.