Using Kolmogorov complexity as input "size"
Consider the parity function (or any other function that depends on all/most bits of the input). For the parity function, $T(w) = \Theta(|w|)$. So 
$$f_n = \Theta(n).$$
On the other hand, 
$$f_n^K = \Theta\left(\frac{1}{|I^K(n)|} \sum_{w:K(w) = n} |w|\right) \geq \Omega\left(\frac{1}{2^n} \max_{w:K(w) = n} |w|\right).$$

Note that $K(2^{2^n}) = O(n)$. Thus $$\max_{w:K(w) = n} |w| \geq 2^{2^{\Omega(n)}}$$ and $f_n^K \geq 2^{2^{\Omega(n)}} / 2^n \to \infty$. Similarly, $K(2^{\dots^{2^{2^n}}}) = O(n)$; thus $f_n^K \geq 2^{\dots^{2^{2^{\Omega(n)}}}}/2^n$ “grows very rapidly”. Moreover, it's not hard to see that there is no computable upper bound for $f_n^K$.