Every sufficiently large positive integer is the average of $n$ distinct primes for certain $n \geq 2$?

For $n$ large enough Vinogradov's theorem gives $\sim C \frac{n^2}{\ln^3 n}$ solutions to $2n+1 = p_1+p_2+p_3$ which gives $\sim B \frac{n^3}{\ln^4 n}$ solutions to $2n = p_1+p_2+p_3+p_4$.

And the same method as Vinogradov gives $\sim A \frac{n^2}{\ln^3 n}$ solutions to $2n = 2q_1+q_2+q_3$.

Thus we have $\sim B \frac{n^3}{\ln^4 n}$ solutions for $2n = p_1+p_2+p_3+p_4$ as the sum of distinct primes.

Vinogradov's theorem follows from a strong form of the PNT in arithmetic progressions. I am quite sure for large enough $k$ there is a three line proof to your $2n = \sum_{j=1}^k p_j$ distinct primes problem just from $\pi(x) \sim \frac{x}{\ln x}$.