I am in a rush so I'll give a hint to the new version. I'll flesh it out if you don't figure it out yourself by tomorrow.
Hint: Let $B_\epsilon$ be the set $\{ \limsup \frac{|f_{n_k}|}{\sqrt{n_k \log \log n_k}} > 1 + \epsilon$. We can write $B_0$ as a countable union of sets of the form $B_\epsilon$ for $\epsilon > 0$. Next, we can show that each $B_\epsilon$ for $\epsilon > 0$ has measure 0, using the decay property you assumed.
Edit: the below addresses a previous version of the question
As stated it is false.
Let $X = (0,1)$, $F$ be the Borel $\sigma$-algebra, and $u$ the usual Lebesgue measure.
Let $f_n(x) \equiv n$. Then we have that
$ u\{ |f_n| \geq \lambda\} = \begin{cases} 1 & n \geq \lambda \\ 0 & n < \lambda \end{cases} $
Observe that if $n \geq \lambda$, $e^{-\lambda/n} \geq e^{-n/n} = e^{-1}$. So the required condition is satisfied for any $C > e$.
Now, we have
$ \frac{|f_{n_k}|}{\sqrt{n_k \log \log n_k}} = \sqrt{ \frac{n_k}{\log\log n_k}} \nearrow \infty $
so we have that precisely everywhere
$ \limsup \frac{|f_{n_k}|}{\sqrt{n_k \log \log n_k}} > 1$