For the first question, note that if we assume Dependent Choice then all $\sigma$-additivity arguments can be carried out perfectly. In such setting using the Caratheodory theorem makes perfect sense, and indeed the Lebesgue measure is the completion of the Borel measure.
It was proved by Solovay that $\mathsf{ZF+DC}$ is consistent with "Every set is Lebesgue measurable", relative to an inaccessible cardinal. This shows that assuming large cardinals are not inconsistent, we cannot prove the existence of a non-measurable set (Vitali sets, Bernstein sets, ultrafilters on $\omega$, etc.) without appealing to more than $\mathsf{ZF+DC}$.
Under the assumption of $\mathsf{DC}$ it is almost immediate that there are only continuum Borel sets, but if we agree to remove this assumption then it is consistent that every set is Borel, where the Borel sets are the sets in the $\sigma$-algebra generated by open intervals with rational endpoints. For example in models where the real numbers are a countable union of countable sets; but not only in such models. Do note that in such bizarre models the Borel measure is no longer $\sigma$-additive.
For sets with Borel codes the proof follows as in the usual proof in $\mathsf{ZFC}$. There are only $2^{\aleph_0}$ possible codes, but there are $2^{2^{\aleph_0}}$ subsets of the Cantor set, all of which are codible-Lebesgue measurable.
Lastly, we have a very good definition for the Lebesgue measure, it is simply the completion of the Borel measure. The Borel measure itself is unique, it is the Haar measure of the additive group of the real numbers. The Lebesgue measure is simply the completion of the Borel measure which is also unique by definition. This is similar to the case where the rational numbers could have two non-isomorphic algebraic closures, but there is always a canonical closure. Even if you can find two ways to complete a measure, "the completion" would usually refer to the definable one (adding all subsets of null sets).
Let $\mathcal A$ be an algebra on a set $X$, $\mu_0$ a premeasure on $\mathcal A$,
and $\mathcal M$ the $\sigma$-algebra generated by $\mathcal A$.
Let $\mu^*$ be the Caratheodory outer measure induced by $\mu_0$, $\mu$ the restriction of $\mu^*$ on $\mathcal M$.
Let $\mathcal C$ be the set of all $\mu^*$-measurable sets, $\bar{\mu}$ the restriction of $\mu^*$ on $\mathcal C$.
Suppose $\mu_0$ is $\sigma$-finite, i.e. there exists a sequence $E_n, n = 1, 2, \cdots$ of members of $\mathcal A$ such that $X = \cup_{n=1}^{\infty} E_n$, $\mu_0(E_n) \lt \infty$ for all $n$. I claim that $(X, \mathcal C, \bar{\mu})$ is the completion of $(X, \mathcal M, \mu)$.
Let $\bar{\mathcal M}$ be the completion of $\mathcal M$ with respect to $\mu$.
Since $\mathcal M\subset \mathcal C$, $\bar{\mathcal M}\subset \mathcal C$.
Hence it suffices to prove that $\mathcal C \subset \bar{\mathcal M}$.
Since $\mu_0$ is $\sigma$-finite, it suffices to prove that if $E\in \mathcal C$ and $\bar{\mu}(E) \lt \infty$, then $E\in \bar{\mathcal M}$.
By the definition of $\bar{\mu}$, for each integer $n \ge 1$, there is a sequence $A_j, j= 1, 2, \cdots$ such that $\sum_{j=1}^{\infty} \mu_0(A_j) \lt \bar{\mu}(E) + 1/n$, where $A_j \in \mathcal{A}$, $E \subset \bigcup_{j=1}^\infty A_j.$ Let $F_n = \bigcup_{j=1}^{\infty} A_j$. Then $E \subset F_n$ and $\mu(F_n) \le \sum_{j=1}^{\infty} \mu_0(A_j) \lt \bar{\mu}(E) + 1/n$.
Let $F = \cap_{n=1}^{\infty} F_n$. Then $F\in \mathcal M$, $E \subset F$, and $\bar{\mu}(E) \le \mu(F) \le \mu(F_n) \lt \bar{\mu}(E) + 1/n$ for all $n\ge 1$. Hence $\bar{\mu}(E) = \mu(F)$.
Similarly there exists $G\in \mathcal M$ such that $F - E \subset G$ and $\mu(G) = \bar{\mu}(F - E)$ = 0.
Then $E = (F - G) \cup (E\cap G)$, $F - G \in \mathcal M$, and $E\cap G$ is a subset of the $\mu$-null set $G$. Hence $E \in \bar{\mathcal M}$. This completes the proof.
Now consider the family $\mathcal A$ of finite disjoint unions of intervals of the form $[a, b)$ or $(-\infty, c)$, where $-\infty \lt a\lt b\le \infty$ and $c\in \mathbb R$.
It is elementary and well known that $\mathcal A$ is an algebra on $\mathbb R$ and there is a unique premeasure $\mu_0$ on $\mathcal A$ such that $\mu_0([a, b)) = b - a$ whenever $a$ and $b$ are finite.
Then $\mathcal M$ and $\mathcal C$ defined above are the families of Borel sets and Lebesgue measurable sets in $\mathbb R$ respectively and you get the picture.
Best Answer
The product $\sigma$-algebra $\bigotimes^n\Sigma$ lies strictly between the Borel $\sigma$-algebra $\mathcal{B}(\mathbb{R}^n)$ and the $\sigma$-algebra $\mathfrak{M}$ of Lebesgue-measurable subsets of $\mathbb{R}^n$ (for $n > 1$). Since the latter is already the completion of the Borel $\sigma$-algebra with respect to the Lebesgue measure, it is also the completion of $\bigotimes^n\Sigma$ with respect to the Lebesgue measure.
Since there are Lebesgue-measurable sets that are not Borel sets, the one strict containment is clear.
For the other, let $N\subset \mathbb{R}$ a non-measurable set. Then $N\times \{0\}^{n-1}$ is a Lebesgue null-set in $\mathbb{R}^n$, but does not belong to $\bigotimes^n\Sigma$. To see the latter, note that for $M\in \bigotimes^n\Sigma$, every section $M_y = \{x \in \mathbb{R} : (x,y) \in M\}, \; y \in \mathbb{R}^{n-1}$ arbitrary, lies in $\Sigma$.