In the following I will prove that the product of three measures is associative. This can easily be generalized to a finite number of measures and possibly to countably many measures. As the OP indicates in his/her post scriptum, it all boils down to proving the associativity result for the $\sigma$-algebras involved.
Let $(X_i,\Sigma_i, \mu_i)$ be a measure space, $i = 1,\ 2,\ 3$. In the following I will use the (standard) notations $$\Sigma_i \times \Sigma_j = \{A \times B : A \in \Sigma_i,\ B \in \Sigma_j\}$$ $$\Sigma_i \otimes \Sigma_j = \sigma\big(\Sigma_i \times \Sigma_j\big),$$ where $\sigma(\cdot)$ indicates the $\sigma$-algebra generated by the argument, i.e. the smallest $\sigma$-algebra that contains the argument. We can finally state the result we want to prove
CLAIM: $\Sigma_1 \otimes \Sigma_2 \otimes \Sigma_3 := \sigma(\Sigma_1 \times \Sigma_2 \times \Sigma_3) = \sigma((\Sigma_1 \otimes \Sigma_2) \times \Sigma_3) = \sigma(\Sigma_1 \times (\Sigma_2 \otimes \Sigma_3)).$
Synopsis of the proof: $\Sigma_i \times \Sigma_j \subset \Sigma_i \otimes \Sigma_j$ is a $\pi$-system. Define the "good set" and apply the $\pi-\Lambda$-system Theorem. Then use the minimality of $\sigma(\cdot)$ several times.
PROOF: Notice that $$(\Sigma_1 \otimes \Sigma_2) \times \Sigma_3 = \{A \times B : A \in \Sigma_1\otimes\Sigma_2,\ B \in \Sigma_3\},$$
then fix $B \in \Sigma_3$ and let $$\Lambda = \{A : A \in \Sigma_1 \otimes \Sigma_2,\ A \times B \in \sigma(\Sigma_1 \times \Sigma_2 \times \Sigma_3)\}.$$
Clearly $\Sigma_1 \times \Sigma_2 \subset \Lambda$. Let's prove that $\Lambda$ is a $\Lambda$-system.
- $X_1 \times X_2 \times B \in \Sigma_1 \times \Sigma_2 \times \Sigma_3$ then $X_1 \times X_2 \in \Lambda$.
- Let $A_1,A_2 \in \Lambda$, $A_1 \subset A_2$. We need to show that $A_2 \setminus A_1 \in \Lambda$, i.e. we need to show that $(A_2 \setminus A_1) \times B \in \sigma(\Sigma_1 \times \Sigma_2 \times \Sigma_3)$. This is easy since $\sigma(\Sigma_1 \times \Sigma_2 \times \Sigma_3)$ is a $\sigma$-algebra, indeed $$(A_2 \setminus A_1) \times B = (A_2 \times B) \setminus (A_1 \times B) = (A_2 \times B) \cap (A_1 \times B)^c \in \sigma(\Sigma_1 \times \Sigma_2 \times \Sigma_3).$$
- Let $\{A_i\}$ be an increasing sequence of sets in $\Lambda$. We need to show that $\cup A_i \in \Lambda$. As before, this easily follows from the fact that $\sigma(\Sigma_1 \times \Sigma_2 \times \Sigma_3)$ is a $\sigma$-algebra. Let's write out the details! $$\Big(\bigcup_{i=1}^{\infty}A_i\Big) \times B = \bigcup_{i=1}^{\infty}(A_i \times B) \in \sigma(\Sigma_1 \times \Sigma_2 \times \Sigma_3).$$
We can finally apply the $\pi-\Lambda$-system Theorem to conclude that $\sigma (\Sigma_1 \times \Sigma_2) \subset \Lambda$ and hence $\sigma(\Sigma_1 \times \Sigma_2) \times B \subset \sigma(\Sigma_1 \times \Sigma_2 \times \Sigma_3).$
Since this is true for every $B \in \Sigma_3$ we obviously get that $$(\Sigma_1 \otimes \Sigma_2) \times \Sigma_3 \subset \sigma(\Sigma_1 \times \Sigma_2 \times \Sigma_3),$$ which in turn yields $$\sigma((\Sigma_1 \otimes \Sigma_2) \times \Sigma_3) \subset \sigma(\Sigma_1 \times \Sigma_2 \times \Sigma_3).$$
Notice we can apply the same reasoning to show that $$\sigma(\Sigma_1 \times (\Sigma_2 \otimes \Sigma_3)) \subset \sigma(\Sigma_1 \times \Sigma_2 \times \Sigma_3).$$
The other inclusion is a lot simpler: $\Sigma_1 \times \Sigma_2 \times \Sigma_3 \subset (\Sigma_1 \otimes \Sigma_2) \times \Sigma_3$, so that $$\sigma(\Sigma_1 \times \Sigma_2 \times \Sigma_3) \subset \sigma((\Sigma_1 \otimes \Sigma_2) \times \Sigma_3),$$ and similarly
$$\sigma(\Sigma_1 \times \Sigma_2 \times \Sigma_3) \subset \sigma(\Sigma_1 \times (\Sigma_2 \otimes \Sigma_3)).$$
This proves the claim. $\blacksquare$
What happens to the measures was covered by David C. Ullrich in the comments section above.
I'm not sure if this provides a complete (heh) answer, but here goes:
For the first question in your final paragraph, it's exactly what it's defined to be; I'm not sure what else to really say about that. For your second question in the final paragraph, it depends. If $X$ and $Y$ are $\sigma$-compact metric spaces, then $\mathcal{B}_{X\times Y}=\mathcal{B}_X\otimes\mathcal{B}_Y.$ so if $\mathcal{B}_1$ is the Borel $\sigma$-algebra on $\mathbb{R},$ then $\mathcal{B}_n=\bigotimes_{j=1}^n\mathcal{B}_1,$ where $\mathcal{B}_n$ denotes the Borel $\sigma$-algebra on $\mathbb{R}^n.$ However, this is not true for Lebesgue measure; if $\mathcal{L}_1$ is the Lebesgue $\sigma$-algebra on $\mathbb{R}$, then $\mathcal{L}_n\neq \bigotimes_{j=1}^n\mathcal{L}_1,$ so we need to complete this to get $\mathcal{L}_n$. Also, this completion is the same as completing $\mathcal{B}_n.$ Note that we can define the Lebesgue measure on $\mathbb{R}^n$ in the Caratheodory manner, so this is a way to view the completion.
Best Answer
When thinking about these types of problems, it's often useful to think of the minimality property of the $\sigma$-algebra generated by some set as being a sort of "induction principle":
(Proof is easy: the conditions imply that for $B = \{ S \subseteq X \mid \phi(S) \}$, we have $A \subseteq B$ and $B$ is a $\sigma$-algebra. Therefore, the $\sigma$-algebra generated by $A$ is contained in $B$.)
The way I think about this principle is: you can informally think of the $\sigma$-algebra generated by $A$ as built up "from the bottom" in steps. First, you start off with $\emptyset$ and elements of $A$. Then, you take complements and countable unions of what you have "so far" and add them to the set. If you repeat this "enough times" then what you get is the generated $\sigma$-algebra. (In fact, you can make this precise using transfinite recursion up to step $\omega_1$.) Then, the conditions say that you start off only with sets satisfying your condition $\phi$, and at each step since you're performing operations on sets satisfying $\phi$, the results will also have to satisfy $\phi$.
Now, what we want to prove in the problem at hand is that whenever $D \in M_1 \otimes M_2$ and $C \in M_3$, then $D \times C \in M_1 \otimes M_2 \otimes M_3$. To do this, we prove by "induction" on $D$, using the above induction principle, that this holds for every $C \in M_3$. That is, for $D \subseteq X_1 \times X_2$, $\phi(D)$ is the property: $\forall C \in M_3, D \times C \in M_1 \otimes M_2 \otimes M_3$. Now, once this is set up, it should be straightforward to prove the "inductive step" conditions (3) and (4), as well as the "base case" (1) and the "base case" (2) where $D = A \times B$, $A \in M_1$, $B \in M_2$. Thus, by the induction principle, it will follow that whenever $D \in M_1 \otimes M_2$, $\phi(D)$ which was what we wanted.
(Note: if you want to replace "$\sigma$-algebra generated by $A$" with "$\sigma$-ring generated by $A$", then all you need to do in the induction principle above is to replace (3) with: For every $S_1, S_2 \subseteq X$, if $\phi(S_1)$ and $\phi(S_2)$, then $\phi(S_1 \setminus S_2)$.)