I will try to answer this both in 18th century terms, and in modern dress.
First off, the réduite equation was introduced on p.208 and named on p.213; it is
$$y^6 + py^3 - n^3/27 = 0$$
The coefficients of this equation are $p$ and $-n^3/27$, thus depending only on the coefficients $n,p$ of the original equation $x^3+nx+p=0$. (Lagrange briefly allows an $mx^2$ term, then transforms it away, resulting in primed coefficients; let's ignore that complication.)
Now $n,p$ can be expressed in terms of the roots of the original equation using Vieta's formulas: just expand out
$$(x-a)(x-b)(x-c) = 0$$
expressing $n,p$ as so-called elementary symmetric functions of $a,b,c$.
Since $n,p$ are symmetric functions of $a,b,c$, it follows that the coefficients of the réduite are too. Thus,
$-n^3/27$ will be some messy expression in $a,b,c$, but clearly it will be symmetric in those roots.
Lagrange has shown that $y=(a + \alpha b + \beta c)/3$, so everything in sight can be written as a rational function of the roots $a,b,c$. (We regard the cube roots of unity as known constants). Imagine what we get if we substitute these functions for $y,n,p$ in the réduite: a big messy equation in rational functions of $a,b,c$. If we now permute $a,b,c$, we will get another such equation. But the coefficients of the réduite are symmetric in $a,b,c$, so they don't change. Just $y$ changes, to some new value. Six possible permutations, six values of $y$, all roots of the same equation. Thus the réduite must have degree 6.
Now let's translate this into modern terms. Lagrange works with two sorts of expressions: rational expressions in the coefficients $n,p$ (or $m,n,p$ if we don't remove the $x^2$ term), and rational expressions in the roots $a,b,c$. Nowadays we like to think set-theoretically, so let's say $E$ is the field of all rational expressions in three letters $a,b,c$. Let $F$ be the subfield generated by the elementary symmetric polynomials
$s_1=a+b+c$, $s_2=ab+ac+bc$, $s_3=abc$. Then the polynomial equation
$$f(X) = (X-a)(X-b)(X-c) = X^3 -s_1 X^2 + s_2 X - s_3 = 0$$
has coefficients in $F$ and roots in $E$. It's not hard, with a modicum of field theory, to show that $E$ is the splitting field of $f(X)$ over $F$. Furthermore, the automorphism group of $E$ over $F$ is just the permutation group on the letters $a,b,c$. The value $y$ has this property: applying the six automorphism to it yields six different values. A standard trick in Galois theory is to pick an element $y$ of the splitting field and form the equation
$$g(X) = \prod_{\sigma\in\text{Aut}(E/F)} (X-\sigma y) = 0$$
The coefficents of $g(X)$ then lie in the fixed field. That's exactly what Lagrange is doing here.
In a comment, the OP raises an issue touching on the differences between the modern and 18th century viewpoints. Let $g(Y)$ be the réduite, where $Y$ is a variable; let $y$ be a rational expression in $a,b,c$ for which $g(y)=0$. Now we permute the roots. The coefficients of $g$ don't change, but $y$ becomes $y'$. How do we know that $g(y')=0$?
The modern answer: permuting the roots induces an automorphism of the field $E$ over $F$, call it $\sigma$. So $\sigma g(y)=g(\sigma y)$ because $\sigma$ is an automorphism leaving all the coefficients of $g$ fixed. So $g(y')=g(\sigma y)=\sigma g(y)=\sigma 0=0$.
Lagrange had no explicit notion of the fields $E$ or $F$, still less of an automorphism of $E$ over $F$. But his remarks about the roots entering equally into the coefficients, and being able to exchange the roots at will, show he had an intuitive grasp of the modern concepts.
Lagrange probably pictured $g(y)=0$ as a big messy rational expression on the left, boiling down formally to 0. Since the equation is formal, not depending on the particular values of $a,b,c$, permuting the roots will not destroy the validity of the formal identity.
Finally, I recommend two articles for help with the pre-history of Galois theory (beside Cox's book): "The Development of Galois Theory from Lagrange to Artin", by B. M. Kiernan (Archive for History of Exact Sciences, v.8 no. 1/2 p. 40-154; p.45-55 discuss Lagrange's work), and "Niels Henrik Abel and the theory of equations" by H. K. Sørensen. (I've also heard good things about Edwards' book Galois Theory.)
