If the spectrum $X$ is $r$-connected, then the map $\Sigma^\infty\Omega^\infty X \to X$ is $(2r+2)$-connected.
Here's a sketch: apply the functor $\Omega^\infty$ to get the map of spaces $$ Q(\Omega^\infty X) \to \Omega^\infty X $$ It will be enoguh to identify the connectivity of the latter.
This map has a section so it is surjective on homotopy in all degrees. In particular your map of spectra is surjective on homotopy in all degrees.
It's enough to compute the degree of injectivity. Let $S^j \to Q(\Omega^\infty X)$ be a map, $j \le 2r+1$. Suppose that its image in $\pi_j(\Omega^\infty X)$ is trivial.
By Freudenthal, the map $$ \pi_j(\Omega^\infty X) \to \pi_j(Q(\Omega^\infty X)) $$ is surjective in this degree ($j \le 2r+1$). So our map desuspends to give a map $S^j \to \Omega^\infty X$. But the composite $$ \pi_j(\Omega^\infty X) \to \pi_j(Q(\Omega^\infty X))\to \pi_j(\Omega^\infty X) $$ is the identity. This means that our desuspension is trivial, hence its $k$-fold suspension is as well.