$\newcommand\Ord{\mathit{Ord}}$The edit to the question has changed it enough so I think it deserves its own answer.

I assume that in reflection we have $\forall \vec{x} \in W_\alpha \, (\phi \to \phi^{W_\alpha})$ holds for all ordinals $\alpha$.

The theory with this strong reflection is inconsistent in a strong sense.

Specifically it fails even when you only consider $W_0$.

Indeed assume that $W_0$ reflects all formulae in the language of $\{\in,j_0\}$ then it reflects $\eta=\exists x(x\notin \operatorname{Dom}(j_\alpha))$, so let $p\in W_0$ be witness of $\eta^{W_0}$.

$W_0$ also reflects $\psi(y)=\exists x\in \Ord (y\in V_x)$, let $r$ be witness of $\psi(p)$ in $W_0$.

Lastly reflect $\phi(y)=y\in \Ord\land V_y\subseteq \operatorname{Dom}(j_\alpha)$ to get that $W_0$ thinks that $\phi(r)$, but this is a contradiction to the choice of $r$, $p$ (note that I only chose 1 arbitrary element, $p$, so I didn't use any choice).

$\newcommand\Ord{\mathit{Ord}}$The edit to the question has changed it enough so I think it deserves its own answer.

I assume that in reflection we have $\forall \vec{x} \in W_\alpha \, (\phi \to \phi^{W_\alpha})$ holds for all ordinals $\alpha$.

The theory with this strong reflection is inconsistent in a strong sense.

Specifically it fails even when you only consider $W_0$.

Indeed assume that $W_0$ reflects all formulae in the language of $\{\in,j_0\}$ then it reflects $\eta=\exists x(x\notin \operatorname{Dom}(j_0))$, so let $p\in W_0$ be witness of $\eta^{W_0}$.

$W_0$ also reflects $\psi(y)=\exists x\in \Ord (y\in V_x)$, let $r$ be witness of $\psi(p)$ in $W_0$.

Lastly reflect $\phi(y)=y\in \Ord\land V_y\subseteq \operatorname{Dom}(j_0)$ to get that $W_0$ thinks that $\phi(r)$, but this is a contradiction to the choice of $r$, $p$ (note that I only chose 1 arbitrary element, $p$, so I didn't use any choice).

The edit to the question has changed it enough so I think it deserves it's own answer.

I assume that in reflection we have $\forall \vec{x} \in W_\alpha \, (\phi \to \phi^{W_\alpha})$ holds for all ordinal.

The theory with this strong reflection is inconsistent in a strong sense.

Specifically it fails even when you only consider $W_0$.

Indeed assume that $W_0$ reflects all formulaes in the language of $\{\in,j_0\}$ then it reflects $\eta=\exists x(x\notin \operatorname{Dom}(j_\alpha))$, so let $p\in W_0$ be witness of $\eta^{W_0}$.

$W_0$ also reflects $\psi(y)=\exists x\in Ord (y\in V_x)$, let $r$ be witness of $\psi(p)$ in $W_0$.

Lastly reflect $\phi(y)=y\in Ord\land V_y\subseteq \operatorname{Dom}(j_\alpha)$ to get that $W_0$ thinks that $\phi(r)$, but this is a contradiction to the choice of $r,p$ (note that I only chose 1 arbitrary element, p, so I didn't use any choice)

$\newcommand\Ord{\mathit{Ord}}$The edit to the question has changed it enough so I think it deserves its own answer.

I assume that in reflection we have $\forall \vec{x} \in W_\alpha \, (\phi \to \phi^{W_\alpha})$ holds for all ordinals $\alpha$.

The theory with this strong reflection is inconsistent in a strong sense.

Specifically it fails even when you only consider $W_0$.

Indeed assume that $W_0$ reflects all formulae in the language of $\{\in,j_0\}$ then it reflects $\eta=\exists x(x\notin \operatorname{Dom}(j_\alpha))$, so let $p\in W_0$ be witness of $\eta^{W_0}$.

$W_0$ also reflects $\psi(y)=\exists x\in \Ord (y\in V_x)$, let $r$ be witness of $\psi(p)$ in $W_0$.

Lastly reflect $\phi(y)=y\in \Ord\land V_y\subseteq \operatorname{Dom}(j_\alpha)$ to get that $W_0$ thinks that $\phi(r)$, but this is a contradiction to the choice of $r$, $p$ (note that I only chose 1 arbitrary element, $p$, so I didn't use any choice).