In $\mathrm{AdS}_5$/$\mathrm{CFT}_4$ the AdS radius $R$ is determined in terms of the string length by the gauge theory t'Hooft parameter as follows \begin{equation} \frac{R}{l_{\rm s}} \sim \lambda^{1/4} \end{equation} Consequently large t'Hooft parameter corresponds to dropping derivative corrections to the spacetime effective action.
In the free O(N) model there is no such parameter. What sets the AdS radius in this case?
[Caveat emptor: this is slightly speculative suggestion from a position of relative ignorance.]
There's also another scale in the game in "ordinary" AdS/CFT: while $\lambda$ sets string length, $N$ sets Planck length. Large $N$ suppresses quantum effects, while large $\lambda$ suppresses stringy effects. Stringy (higher derivative) effects have no obvious parallel in Vasiliev, and $\lambda$ has no obvious parallel in the $O(N)$ model. On the other hand, Quantum effects seem like a natural thing to have in the bulk, and $N$ looks rather like the $N$ in SYM. So perhaps the answer is that it's $N$ setting the radius of AdS relative to the Planck length.
Maybe the view of Vasiliev being something like a tensionless, $l_s\to\infty$ limit of strings, with the first Regge trajectory becoming massless, would be a helpful point of view here? I'd need to think a little more to make this feeling any more precise...
