So we solved this today, in a way similar to how the Hilbert basis theorem might be proved. First show the following lemma, then the claim that $D^n(B)$ is artinian as a module over $B[t_1,\dotsc,t_n]$ is a direct consequence.
Lemma If $M$ is an artinian $R$-module, then $D^1 \otimes M$ is an artinian $R[t]$ module.
Proof: We can think of elements in $D^1 \otimes M$ as polynomials with coefficients in $M$. So each element $f$ has a degree $\deg f$ and a leading coefficient $lc(f)$. Now consider a descending chain $$ N_1 \supset N_2 \supset \dots$$ an set $M_{k, l } := \{ lc(f) : f \in N_k, \deg(f) = l \} \subset M$. Clearly $M_{k,l}\supset M_{k+1,l}$ and $M_{k,l}\supset M_{k,l+1}$.
Now observe that $N_k = N_{k+1}$ if and only if $M_{k,l} = M_{k+1,l}$ for all $l$. "$\Rightarrow$" is clear, so for "$\Leftarrow$" consider $f \in N_k \setminus N_{k+1}$ with minimal degree. Then there is some $g \in N_{k+1}$ with $lc(f) = lc(g)$ and $\deg(f) = \deg(g)$. So $f - g \in N_k\setminus N_{k+1}$ has lower degree, which is a contradiction.
Next consider the diagram \begin{align*} \dots && \dots && \dots \\ \cap && \cap && \cap\\ M_{3,1} &\supset& M_{3,2} &\supset& M_{3,3} &\supset& \dots\\ \cap && \cap && \cap\\ M_{2,1} &\supset& M_{2,2} &\supset& M_{2,3} &\supset& \dots\\ \cap && \cap && \cap \\ M_{1,1} &\supset& M_{1,2} &\supset& M_{1,3} &\supset& \dots \end{align*} Now the sequence $M_{1,1} \supset M_{2,2} \supset M_{3,3} \supset \dots$ has to stabilize at some point, because $M$ is artinian, say at $M_{k,k}$. Now it is easy to see that $M_{n,m} = M_{k,k}$ for all $n,m\geq k$. Thus there are only finitely many rows in the diagram which are not yet equal, but they will equal after finitely many steps. So the criterion above applies and we see that the sequence $N_k$ has to stabilize, proving the claim.
This directly proves the following
Corollary If $B$ is artinian, then $D^n(B)$ is artinian as a module over $B[t_1,\dotsc,t_n]$.
Proof: Apply the lemma inductively.