Do Arithmetic Mean and Geometric Mean of Prime Numbers converge?

Your conjecture for GM was proved in 2011 in the short paper On a limit involving the product of prime numbers by József Sándor and Antoine Verroken.

Abstract. Let $p_k$ denote the $k$th prime number. The aim of this note is to prove that the limit of the sequence $(p_n / \sqrt[n]{p_1 \cdots p_n})$ is $e$.

The authors obtain the result based on the prime number theorem, i.e., $$p_n \approx n \log n \quad \textrm{as} \ n \to \infty$$ as well as an inequality with Chebyshev's function $$\theta(x) = \sum_{p \le x}\log p$$ where $p$ are primes less than $x$.

We can use the simple version of the prime counting function $$p_n \approx n \log n$$ and plug it into your expressions. For the arithmetic one, this becomes $$\lim_{n \to \infty} \frac {\sum_{i=1}^n p_i}{np_n}=\lim_{n \to \infty} \frac {\sum_{i=1}^n i\log(i)}{np_n}=\lim_{n \to \infty} \frac {\sum_{i=1}^n \log(i^i)}{np_n}\\=\lim_{n \to \infty} \frac {\log\prod_{i=1}^n i^i}{np_n}=\lim_{n \to \infty}\frac {\log(H(n))}{n^2\log(n)}$$ Where $H(n)$ is the hyperfactorial function. We can use the expansion given on the Mathworld page to get $$\log H(n)\approx \log A -\frac {n^2}4+\left(\frac {n(n+1)}2+\frac 1{12}\right)\log (n)$$ and the limit is duly $\frac 12$

I didn't find a nice expression for the product of the primes.

Here is a general answer to this which will solve the case for AM, GM and HM in one shot.

Observe that since $p_n \sim n\log n$, as $n \to \infty$ the proportion of numbers formed by the sequence of ratios $\frac{p_1}{p_n},\frac{p_2}{p_n} \ldots, \frac{p_{n-1}}{p_n}$ which fall inside any sub-interval within $(0,1)$ is proportion to the length of that interval i.e. the sequence $\frac{p_r}{p_n}$ approaches equidistributed in $(0,1)$ [for the proof of equi/uniform distribution, see the comment below by Mariuslp]. Hence, for an equidistributes sequence, we have:

Theorem: Let $p_k$ be the $k$-th prime and let $f$ be a continuous function Riemann integrable in $(0,1)$ then,

$$ \lim_{n \to \infty}\frac{1}{n}\sum_{r = 1}^{n}f\Big(\frac{p_r}{p_n}\Big) = \int_{0}^{1}f(x)dx. $$

(See the proof in this MO link for a direct proof). Taking $f(x) = x, \log(x)$ and $\frac{1}{x}$ respectively with some manipulations, we get the required limit for AM, GM and HM as $\frac{1}{2},\frac{1}{e}$ and $0$ respectively.

Example: Showing the case for GM due to request in the bellow comments. Let $$ \lim_{n \to \infty}\frac{(p_1 p_2 \ldots p_n)^{1/n}}{p_n} = \lim_{n \to \infty}\Big(\frac{p_1}{p_n}\Big)^{1/n} \Big(\frac{p_2}{p_n}\Big)^{1/n} \ldots \Big(\frac{p_n}{p_n}\Big)^{1/n} = l $$

Clearly, $0 < l < 1$. Taking logarithm on both sides, we have $$ \log l = \lim_{n \to \infty} \frac{1}{n} \sum_{r = 1}^{n}\log \Big(\frac{p_r}{p_n}\Big) = \int_{0}^{1} \log x dx = -1. $$

Hence $l = 1/e$.