I am interested in learning about the homotopy groups of the spectrum $D(n)$ at the prime $2$ which is defined as the cofibre of the diagonal map
$$Sp^{2^{n-1}}S^0 \to Sp^{2^n}S^0$$
where $Sp$ is the symmetric power functor. In particular, I am interested in the spectrum $D(2)$. Indeed, I imagine that there would not be a complete description of these groups as the $D(0)=S^0$ and $D(1)=\Sigma P_{-1}$ tells me that even the cases of $n=0,1$ is quit difficult. But, I am still wondering if there exists any computations, at least in a range, on homotopy groups of these spectra. In particular, does the equality
$$\mathrm{hocolim}\ D(n)=H\mathbb{Z}/2\ (*)$$
suffice to show that for any $n$ then exist $b_n$ so that $\pi_iD(n)=0$ for $i>b_n$?! I would be very grateful if someone can point out at a reference with some state of art computations on (homotopy groups of) these spectra. Thank you very much in advanced.
ADDED By work of Mitchell and Priddy we know tht $H^*D(n)$ has a basis consisting of classes $Sq^Iu_n$ with $I$ being an admissible sequence of length not more than $n$ and $u_n\in H^0D(n)$ is a generaor. In particular, I think, this tells me that $\pi_0D(n)=0$ if $n>0$ and the integer $c_n$ in the comments below will not exist unless we assume $c_n>0$. What I read off from $(*)$ is that for any $i>0$ there exists a $d_i>0$ so that $\pi_iD(n)=0$ for $n>d_i$.
Further Addendum Thanks to the comments below, I noticed that the kernel of the map $A=H^*H\mathbb{Z}/2\to H^*D(n)$ induced by $i_n:D(n)\to H\mathbb{Z}$ has a basis of all classes $Sq^I$ with $I$ admissible and of length at least $n+1$. Here, $I=(i_1,\ldots,i_{n+1})$ is admissible if $i_j\geqslant 2i_{j+1}$. I think the admissible sequence $I$ of length $n+1$ and least dimension would be $I=(2^n,2^{n-1},\ldots,2,1)$ whose dimension is $\sum_{i=0}^{n}2^i=2^{n+1}-1$. Therefore, I conclude that $i^*$ is an isomorphism in $H^*$ for $*\leqslant 2^{n+1}-2$. Hence, if $n>0$ then $$i_*:\pi_*D(n)\to \pi_*H\mathbb{Z}/2$$ is an isomorphism for $*\leqslant 2^{n+1}-2$. Does it sound reasonable?