Finding a $g(n)$ so that $f(n) = O(g(n))$
For any $f$ we have $f \in O(f)$ i.e. trivial answer to your question is $g=f$.  Presumably  you need  to clarify more exactly conditions on $g$. On other hand if we change all members of sum with maximum one, then we have $f \in O(n^{28})$.
Now about questions:

$c$ is independent from $n$ by definition
outgoing from 1. you cannot consider $c = (n-1)n^{5}$