Chen's theorem, that every sufficiently large even number is the sum of a prime and another number with at most two prime factors, is a result in this direction. That is, taking $n=2$ in your expression accounts for those even numbers which are a sum of two (odd) primes, and taking $n=p+1$ where an even number is a sum of a prime and a product $pq$ of two odd primes (semiprime or "almost prime") accounts for the others.
Some work has been done to improve on Chen's work, but I don't know of an explicit bound beyond which his sieve methods show this representation holds.
Edit: The construction above almost meets the bound $n^2 < C = 2m$ if we choose $p$ to be the lesser prime factor of the semiprime summand (if applicable). Unless the factors are equal (the semiprime is a square), $(p+1)^2 < C$.
Without the restriction on $n$ I don't think your result is difficult to prove for $C \gt 4$. Consider the so-called Bertrand's postulate that a prime exists between $m$ and even number $2m$. Let $q$ be the least such prime, so that $2m-q$ is an odd number. If this is prime, we are done (as explained above for $n=2$ in your notation). If this is an odd number greater than $3$, then is has an odd prime factor $p$, and again we are done as before ($2m = ((2m-q)/p)*p + q$ and the factor $(2m-q)/p$ is (being odd) one less than an even $n$).
The only difficulty I can see is with very small even numbers. That is, we cannot express $2$ or $4$ as sums of two (positive) odd primes, nor can we express them as a sum of the kind you ask about (if we restrict ourselves to positive summands).