No. At least for $w_1,w_2$ polylog$(n)$ and $w_3 \geq n^{\epsilon}$ for some $\epsilon \in \Omega(1)$, there is no such graph.

Suppose otherwise. Pick the set $B$ of $b = 2w_1w_2\log n$ vertices of the highest degree in $G$. The sums of the degrees of these vertices has to be at least $\frac{bD}{2w_2} \geq n \log n$.

However, let for each vertex $u \in G$, let $N_B(u)$ be the number of neighbors that $u$ has in $B$. Then $\sum_{u \in V(G)} N_B(u)$ is the sums of the degrees of vertices in $B$ which is $\frac{bD}{2w_2}$ $\geq n \log n$.

Furthermore, $N_B(u)$ is upper-bounded by $b$. So there are at least
$(\frac{bD}{2w_2} - n)/b$ vertices $u$ adjacent to 2 or more vertices in $B$.
As $\frac{bD}{2w_2}$ is at least $n \log n$, it follows that $\frac{bD}{2w_2}-n$ is at least $\frac{bD}{4w_2}$, and so (*) there are at at least
$\frac{D}{4w_2}$ vertices $u$ adjacent to 2 or more vertices in $B$

However, the bound on the max codegree implies that (**) the number of vertices $u$ adjacent to 2 or more vertices in $B$ is no more than $\frac{b^2D}{w_3}$ which is $O(\frac{b^2D}{n^{\epsilon}})$ if $w_3 = n^{\epsilon}$.

However, note that $b$ is only polylog if $w_1$ and $w_2$ are polylog, so at most only one of (*) and (**) may hold if both $w_1$ and $w_2$ are polylog and $w_3$ is $\Omega(n^{\epsilon})$ for some $\epsilon \in \Omega(1)$.