Claim: The $\mathrm{pcf}$ structure on $(\aleph_1 \times \aleph_2 \times \dots) \times (\aleph_1 \times \aleph_2 \times \dots) \times \dots$ gives nothing new.
Proof: For notational convenience, let $A : \omega \to \mathrm{Reg}$ be defined by $A(n) = \aleph_n$ and let $B : \omega \cdot \omega \to \mathrm{Reg}$ be defined by $B(\omega\cdot m + n) = \aleph_n$. Define
$$\mathrm{pcf}(A) = \{\mathrm{cf}(\Pi_{n<\omega}A(n)/U)\ :\ U \in \beta \omega\}$$$$\mathrm{pcf}(B) = \{\mathrm{cf}(\Pi_{\alpha<\omega\cdot\omega}B(\alpha)/U)\ :\ U \in \beta (\omega\cdot\omega)\}$$
Where $\beta X$ denotes the set of all ultrafilters on $X$. It's not hard to see that $\mathrm{pcf}(A) \subseteq \mathrm{pcf}(B)$ and since $\mathrm{pcf}(A)$ is an interval of regular cardinals, it suffices to show that $\max \mathrm{pcf}(B) = \max \mathrm{pcf}(B)$. We know that we can find an everywhere-pointwise-dominating family on $\Pi A$ of size $\lambda := \max\mathrm{pcf}(A)$. If we can find a dominating family on $\Pi B$ of that same size, we'll be done.
So, given a dominating family $\mathcal{F}$ on $\Pi A$ of size $\lambda$, just let $\mathcal{F}^\ast \subseteq \Pi B$ consist of functions of the form:$$f^\ast (\omega\cdot m + n) = f(n)$$for each $f \in \mathcal{F}$. Now given $g \in \Pi B$, define:$$g'(\omega\cdot m + n) = \sup_{m' \in \omega}g(\omega\cdot m' + n)$$We see that $g' \geq g$ everywhere pointwise, and $g' \in \Pi B$ since we're always taking countable suprema within uncountable regular cardinals. Since $g'$ has the same value at any of its coordinates that correspond to the same $\aleph_n$, it's clear that there's some $f \in \mathcal{F}$ such that $f^\ast \in \mathcal{F}^\ast$ dominates $g'$ everywhere, and hence $g$ everywhere.
