I'm quite late answering this question, but I figured it still deserves an answer. The answer to your question is "no": when $d,d^\prime \geq 3$ there does not exist such a finite (or even countable) set of positive maps.

In arXiv:1209.0437, we considered the following problem: given a set of positive maps $\mathcal{Q}$ from $\mathcal{M}_d$ to $\mathcal{M}_{d^\prime}$, what is the set

$$\mathcal{C}_\mathcal{Q} \stackrel{\text{def}}{=} \{\sum_i \psi_i \circ P_i \circ \phi_i : P_i \in \mathcal{Q}, \psi_i \text{ and } \phi_i \text{ are completely positive for all $i$}\}?$$

As you noted, Størmer and Woronowicz showed that if $\mathcal{Q} = \{T\}$ (where $T$ is the transpose map) and $dd^\prime \leq 6$, then $\mathcal{C}_\mathcal{Q}$ is the set of all positive maps.

We didn't prove it in the paper, but as an offshoot Łukasz Skowronek proved in the $d,d^\prime\geq 3$ case that if $\mathcal{C}_\mathcal{Q}$ is the set of all positive maps then $\mathcal{Q}$ must be uncountable. I don't believe he ever did end up formally writing up the result, but he at least gave a talk about it (slides available here).

The slides go through the proof in pretty good detail: the rough idea is that there is a known uncountably infinite set of positive maps on $\mathcal{M}_3$ that are each exposed in the set of positive maps and are all indecomposable (i.e., can not be written in the form $\phi_1 + T \circ \phi_2$ for some completely positive $\phi_1,\phi_2$) due to Ha and Kye (see arXiv:1108.0130). Łukasz showed that there is no countable set $\mathcal{Q}$ such that $\mathcal{C}_\mathcal{Q}$ contains even just this particular (relatively small) set of positive maps.

**Update [May 23, 2016]:** It seems that Łukasz finally got around to formally writing up this result, and it is now proved explicitly in this new paper).

positive map, as well as ofcompletely positive map. – Denis Serre Feb 16 '11 at 9:29positiveif sends the positive cone to the positive cone. It is$k$-positiveif $\Phi$ tensored with the identity map on $k \times k$-matrices is positive. It iscompletely positiveif is $k$-positive for all $k > 0$. – Michael Feb 16 '11 at 10:00