Cofiber of the inclusion of an $E_0$-algebra $M$ into the free $E_k$-algebra generated by it
I don't have a general answer but I can confirm your intuition in characteristic $0$. Then we can work with chain complexes and $E_k$-algebras are Poisson $k$-algebras for $k\geq 2$, i.e. commutative DGAs equipped with a compatible Lie bracket of degree $k-1$.
For the sake of simplicity let us consider an $M$ where the unit map splits, $M=A\oplus N$. We can assume that the complex $N$ is trivial in degrees $<d$ (you can even suppose that $N$ has trivial differential if this helps). Then $F(M)$ can be described as follows,
\[ \begin{array}{rcl} F(M)&=&A\oplus S(\Sigma^{1-k}Lie(\Sigma^{k-1}N))\cr &=&A\oplus \bigoplus_{m\geq 1}(\bigoplus_{n\geq 1}\Sigma^{1-k}Lie(n)\otimes_{\Sigma_n} (\Sigma^{k-1}N)^{\otimes^n})^{\otimes^m}_{\Sigma_m} \end{array} \]
Here $S(-)$ and $Lie(-)$ in the first line are the free commutative and Lie algebra functors, and $Lie(n)$ in the second line denotes the Lie operad, which is concentrated in degree $0$.
The map $M\rightarrow F(M)$ is the inclusion of the factor $A$ and the factor obtained for $m=1$ and $n=1$, hence the cofiber is simply obtained by removing these two direct summands.
The complex $\Sigma^{k-1}N$ is $d+k-1$ connected, hence its $n^{\text{th}}$ tensor power is $n(d+k-1)$ connected. Tensoring with $Lie(n)$ doesn't affect connectivity, and if we desuspend $k-1$ times it is $n(d+k-1)-(k-1)$ connected, which is $\geq 2d$ for $n\geq 2$. For $n=1$ the tensor power on $m$ is at least $m\geq 2$, so we get again connectivity degree $\geq 2d$. Note that coinvariants do not reduce connectivity.
Let me try to turn my comment into an answer. Again, the intuition is to filter $\mathrm{Ind}_{\mathbb{E}_0}^{\mathbb{E}_k}M$ by arity, and the result (at least in slightly less generality) is certainly classical, but I'll give a proof in the language you use above. I will also make the following assumptions (which hopefully you are making too):
- The $\mathbb{E}_{k+1}$-ring $A$ is connective.
- Let $\overline{M}$ denote the cofiber of the unit $A \to M$. Then $\overline{M}$ is $d$-connective (i.e. $(d-1)$-connected).
Proof: I claim that there is a filtration $\{F_k\mathrm{Ind}_{\mathbb{E}_0}^{\mathbb{E}_k}M\}$ of the induced algebra such that
- The associated graded is $\bigoplus_{n\ge 0} \mathrm{Sym}^n_{\mathbb{E}_k}(\overline{M})$
- $F_1=M$ and the map $F_1=M \to \mathrm{Ind}_{\mathbb{E}_0}^{\mathbb{E}_k}M$ is the standard one.
I'll build that in a minute but first let's check that it proves the result. It suffices to check that $\mathrm{Sym}^n_{\mathbb{E}_k}(\overline{M})$ is at least $2d$-connective as soon as $n\ge 2$. But, by definition, this object is a colimit over a diagram whose objects all look like $\overline{M}^{\otimes_An}$. Since the $t$-structure is compatible with the tensor product (since $A$ is connective), these are $nd$-connective, and the subcategory of $nd$-connective objects in closed under colimits (since connectivity is checked on underlying spectra and it's true there). So the result is proved assuming we have the filtration. Now let's build it.
First recall the construction of $\mathrm{Ind}_{\mathbb{E}_0}^{\mathbb{E}_k}M$. We're meant to consider all the ways to build tensor powers like $M^{\otimes_An}$ (one for each rectilinear embedding $\coprod_n I^k \hookrightarrow I$) and then glue things together via the various maps $M^{\otimes_An} \to M^{\otimes_Am}$ induced by the unit when $n\le m$. Explicitly, let $D$ denote the $\infty$-category whose objects are rectilinear embeddings $\coprod_n I^k \hookrightarrow I$ ($n$ arbitrary) and whose morphisms are embeddings $\coprod_n I^k \hookrightarrow \coprod_m I^k$ compatible with the structure map up to isotopy and which are injective on path components. (A rigorous definition is the fiber product in HA.3.1.3.1 in the case $\mathcal{C} = \mathsf{LMod}_A$, $\mathcal{A}^{\otimes} = \mathbb{E}_0^{\otimes}$, and $\mathcal{B}^{\otimes} = \mathbb{E}_k^{\otimes}$). Then $M$ extends to a diagram $M^{(-)}: D \to \mathsf{LMod}_A$ and the module underlying $\mathrm{Ind}_{\mathbb{E}_0}^{\mathbb{E}_k}M$ is the colimit of this diagram.
Now, $D$ admits an evident filtration by $n$, $D_{\le n} \subset D_{\le n+1}$. Restricting and taking colimits defines a filtration $\{F_k\mathrm{Ind}_{\mathbb{E}_0}^{\mathbb{E}_k}M\}$ and I want to show it has the two properties above. The second should be straightforward from the definition, so I'll just do the first.
Notice that there is a natural map $F_n/F_{n-1} \to \mathrm{Sym}^n_{\mathbb{E}_k}(\overline{M})$ coming from the natural maps $M^{\otimes_An} \to \overline{M}^{\otimes_An}$. We want this to be an equivalence. We'll prove this by induction on $n$. When $n=1$ the result is true by inspection: we're looking at the map $F_0=A \to F_1=M$ and taking the cofiber to get $\overline{M}$. I think one can prove the general case by induction on $n$, but I would have to come back and write the details if you want them.
The following is an expansion of Dylan's suggestion to use an arity filtration. The reader's digest version is: there's a filtration (which I'm going to construct below) whose associated graded is the free $E_k$-algebra on the cofiber $M/A$, this free algebra has a decomposition into homotopy colimits of diagrams of terms $(M/A)^{\otimes_A n}$, and the "new" terms in $F(M)$ are the quadratic and higher terms with $n > 1$ whose connectivity is roughly $n$ times the connectivity of $M/A$.
Let $\widetilde F$ be the simpler adjunction between left $A$-modules and $E_k$-algebras. We have $F(A \oplus X) \simeq \widetilde F(X)$ naturally. That helps a little, because $\widetilde F$ has a simpler decomposition into the "symmetric power" terms: $$ \widetilde F(X) = \bigoplus_{n \geq 0} {\rm Sym}^n_{E_k}(X) $$ Moreover, any map $A \oplus X \to A \oplus Y$ induces a map $\widetilde F(X) \to \widetilde F(Y)$ that only decreases symmetric power filtration, because the terms involving $A$ all become unit terms. Thus, there's a natural filtration objects of the form $F(A \oplus X)$ whose associated graded is $\widetilde F(X)$.
Let's take our general $E_0$-algebra $A \to M$ now and write it as the homotopy colimit of a simplicial object: $$ M \simeq {\rm hocolim} \{ A \oplus M \leftleftarrows A \oplus A \oplus M \dots\} $$ which is basically saying that $M$ is equivalent to the mapping cylinder of $A \to M$. Then we can apply $F$ to this because $F$ commutes with sifted colimits: $$ F(M) \simeq {\rm hocolim} \{ F(A \oplus M) \leftleftarrows F(A \oplus A \oplus M) \dots \} = {\rm hocolim} \{\widetilde F(M) \leftleftarrows \widetilde F(A \oplus M) \dots\} $$ Applying our natural filtration, we find that there's a filtration of $F(M)$ whose associated graded has terms $$ {\rm hocolim} \{{\rm Sym}^n_{E_k}(M) \leftleftarrows {\rm Sym}^n_{E_k}(A \oplus M) \dots \} \simeq {\rm Sym}^n_{E_k} (M/A) $$ because ${\rm Sym}^n_{E_k}$ also commutes with sifted hocolims.
In this filtration, the bottom two graded terms are ${\rm Sym^0}(M/A) = A$ and ${\rm Sym^1}(M/A) = M/A$, representing the map $M \to F(M)$ in the associated graded. Thus there is a filtration of the cofiber of $M \to F(M)$ whose associated graded is $$ \bigoplus_{n \geq 2} {\rm Sym}^n_{E_k} (M/A). $$
Great. So how do we get connectivity estimates on these functors ${\rm Sym}^n_{E_k}$?
The functor ${\rm Sym}^n_{E_k}$ has the following description. Let $E_k(n)$ be the $n'th$ space in the $E_k$-operad, and $B_k(n) = {\rm hocolim}_{\Sigma_n} E_k(n)$ be the quotient by the action of the symmetric group. This space is homotopy equivalent to the configuration space of $k$ points in an $n$-disk.
The fact that ${\cal C}$ is $E_k$-monoidal means, in particular, that there's a map $$ B_k(n) \to {\rm Fun}({\cal C}, {\cal C}) $$ from this configuration space to the space of functors ${\cal C} \to {\cal C}$. We get this by starting with a $\Sigma_n$-equivariant map $E_k(n) \to {\rm Fun({\cal C}^n, {\cal C})}$ determined by the monoidal structure, following it with the diagonal ${\cal C} \to {\cal C}^n$, and observing that the target ${\rm Fun}({\cal C}, {\cal C})$ has trivial $\Sigma_n$-action so we can pass to the quotient.
Said another way: the points $\alpha$ of this configuration space parametrize $n$-fold tensor power functors $X \mapsto X^{\otimes_\alpha n}$. All of these are equivalent to $X \mapsto X^{\otimes_A n}$, but there's still a nontrivial space of them. We then have $$ {\rm Sym}^n_{E_k}(X) \simeq {\rm hocolim}_{\alpha \in B_k(n)} (M/A)^{\otimes_\alpha n}. $$ Hoewever, hocolims preserve connectivity and so we only need a lower bound on the connectivity of the inside term, which is equivalent to $(M/A)^{\otimes_A n}$.
As a result (assuming $A$ is connective), we find that if $M/A$ is $d$-connected then $(M/A)^{\otimes_\alpha n}$ is $n(d+1)-1$-connected, and thus the whole sum is at least $(2d+1)$-connected.