reference request - nth-order generalizations of the arithmetic-geometric mean

You want Maclaurin's inequality. Given $n$ positive numbers $a_1, a_2,dots,a_n$,
write
$$
(x+a_1)(x+a_2)cdots(x+a_n) = x^n + S_1x^{n-1} + cdots + S_{n-1}x + S_n,
$$
so $S_i$ is the $i$-th elementary symmetric function of $a_1,dots,a_n$.
For $i = 1,dots,n$, set $A_i = S_i/binom{n}{i}$. When
$n = 2$, $A_1 = (a_1+a_2)/2$ and $A_2 = a_1a_2$. Maclaurin's inequality is that
$$
A_1 geq sqrt{A_2} geq sqrt[3]{A_3} geq cdots geq sqrt[n]{A_n},
$$
where the inequality signs are all strict unless $a_1,dots,a_n$ are all equal.
The inequality of the outer terms, $A_1 geq sqrt[n]{A_n}$, is the arithmetic-geometric mean inequality for $n$ positive numbers.

From a list of $n$ positive numbers $a_1,dots,a_n$ we have produced another list of $n$ positive numbers $A_1,sqrt{A_2},dots,sqrt[n]{A_n}$. The construction can be repeated.

Theorem: All the terms in the list tend to the same limit.

Off the top of my head I can't recall a reference where this is proved.
It was studied by Meissel in 1875 for $n = 3$.

For example, if we start with the three numbers 1, 2, 3 then
after 4 iterations the three numbers we get all look like 1.9099262335 to 10 digits after the decimal point.

[Edit: Here is a proof of the common limit, based on Will Sawin's first comment below to my answer. Order the numbers $a_1,dots,a_n$ so that $a_1 geq cdots geq a_n > 0$. By Maclaurin's inequality (or really just the arithmetic-geometric mean inequality)
$A_1 geq sqrt[n]{A_n}$ and we will bound
$A_1 - sqrt[n]{A_n}$ from above in terms of $a_1 - a_n$ by bounding $A_1$ from above and $sqrt[n]{A_n}$ from below using just $a_1$ and $a_n$. To bound $A_1$ from above,
$$
A_1 = frac{a_1 + cdots + a_n}{n} leq frac{(n-1)a_1 + a_n}{n} = a_1 - frac{a_1 - a_n}{n}
$$
and to bound $A_n$ from below we write $A_n = a_1cdots a_n geq a_n^n$, so
$$
sqrt[n]{A_n} geq a_n.
$$
Therefore
$$
0 leq A_1 - sqrt[n]{A_n} leq left(a_1 - frac{a_1 - a_n}{n}right) - a_n = left(1 - frac{1}{n}right)(a_1 - a_n).
$$
Start from an $n$-tuple $(a_1,a_2,dots,a_n)$ which is ordered so that $a_1 geq cdots geq a_n > 0$ and construct the $n$-tuple $(A_1,sqrt{A_2},dots,sqrt[n]{A_n})$ and keep repeating this,
which produces a sequence of $n$-tuples $(a_1^{(k)},a_2^{(k)},dots,a_n^{(k)})$ for $k = 0,1,2,dots$, where $a_i^{(0)} = a_i$. Let's look at the sequence of first coordinates $a_1^{(k)}$. An arithmetic mean of positive numbers is bounded above by the largest number, so $a_1 = a_1^{(0)} geq a_1^{(1)} geq a_1^{(2)} geq cdots > 0$. Therefore
the sequence $a_1^{(k)}$ converges as $k rightarrow infty$. (The limit is positive because the sequence of last coordinates $a_n^{(k)}$ is non-decreasing and $a_1^{(k)} geq a_n^{(k)} geq a_n^{(0)} = a_n$ for all $k$.) The above calculation shows
$$
0 leq a_1^{(k)} - a_n^{(k)} leq left(1 - frac{1}{n}right)(a_1^{(k-1)} - a_n^{(k-1)}),
$$
so $0 leq a_1^{(k)} - a_n^{(k)} leq (1 - 1/n)^k(a_1 - a_n)$. Letting $k rightarrow infty$ we see the sequence of last coordinates $a_n^{(0)},a_n^{(1)},a_n^{(2)},dots$ converges to the limit of the sequence of first coordinates $a_1^{(0)}, a_1^{(1)}, a_1^{(2)},dots$. Since $a_1^{(k)} geq a_i^{(k)} geq a_n^{(k)}$, each intermediate sequence $a_i^{(0)},a_i^{(1)},a_i^{(2)},dots$ converges to the same limit.
]