We know that for 3 variables $(A=0,B=1,C=1)$, $f_1 = (A \mathop{\text{ XNOR }} B \mathop{\text{ XNOR }} C) = 1$, since the input has even number of $1$'s.
But if we were to do this step by step, $f_2 = (A \mathop{\text{ XNOR }} (B \mathop{\text{ XNOR }} C)) = (A \mathop{\text{ XNOR }} (1 \mathop{\text{ XNOR }} 1))$
Now, $1 \mathop{\text{ XNOR }} 1 = 1$. So, $f_2 = (A \mathop{\text{ XNOR }} 1) = 0 \mathop{\text{ XNOR }} 1 = 0$.
Why is $f_1 \neq f_2$ ?
There is no such rule as you mention, that an expression with an even number of $1$ containing only XNOR will evaluate to $1$. Your evaluation in the second part is correct. Thus there is not contradiction, just a faulty assumption/theorem from your part.