A "universally non Hypercomplete" $\infty$-topos via Goodwillie calculus? My question is :
Is there a classifying $\infty$-topos for $\infty$-connected objects ? Does this $\infty$-topos has a nice description (as an $\infty$-category ) ?
What I mean by $\infty$-connected object is an object $X$ such that all the $\pi_n(X)$ (seen as truncaturated objects of the $\infty$-topos) are isomorphic to the terminal object.
And by classifying $\infty$-topos I mean that for any other $\infty$-topos $\mathcal{E}$ there is a (natural) equivalence of $\infty$-category between $\infty$-connected object in $\mathcal{E}$ and geometric morphisms from $\mathcal{E}$ to our topos $\mathcal{T}$, this (inverse of this) equivalence being induced by $f \mapsto f^* (X)$ where $X$ is a given ("universal") $\infty$-connected object of $\mathcal{T}$.
I am relatively convinced that such an $\infty$-topos exists, so I'm more interested in knowing if it has an interesting description (something simpler than than a localization of the category of simplicial presheaves over the category of finite simplicial set would be great ! ).
This topos should have a unique point which should be its hypercompletion, I know one $\infty$-topos with this property (mentioned for example in this answer ). I guess it would be way too beautiful if this was the answer, but there is at least be a geometric morphism between them...
 A: To summarize the discussion in the comments:


*

*$Fun(\mathsf{FinTop},\mathsf{Top})$ classifies objects
(Proof: since a left adjoint functor $Fun(\mathsf{FinTop},\mathsf{Top}) \to \mathcal E$ corresponds to an arbitrary functor $\mathsf{FinTop}^\mathrm{op} \to \mathcal E$, and if that functor is left exact, it just corresponds to a functor from the terminal category to $\mathcal E$.)

*$\infty$-connected objects are objects with a certain property, so they are classified by a localization $\mathcal T$ of $Fun(\mathsf{FinTop},\mathsf{Top})$,  which is precisely the full subcategory of $\infty$-connected objects  [$Fun(\mathsf{FinTop},\mathsf{Top})$ is a presheaf topos, hence hypercomplete. So it's a bit more subtle.].

*Jacob Lurie's comment describes this localization as being at the Grothendieck topology generated by all maps ( but I don't know why this is so  This is explained in Denis Nardin's answer).

*This localization can be computed using the fact that every object $X$ has a finest cover in this topology, the map in $\mathsf{FinTop}^\mathrm{op}$ corresponding to the map $X \to \ast$ in $\mathsf{FinTop}$. We take a limit over the Cech nerve of this singleton cover in $\mathsf{FinTop}^\mathrm{op}$, whose $n$th level consists of an "$n$-pointed cone on $X$".

*These cones also show up in Goodwillie calculus, so we see that an $n$-excisive functor $F$ lies in $\mathcal T$ because the descent object stabilizes at the $n$th stage. Moreover, the localization is closed under limits, so it contains all limits of $n$-excisive functors for varying $n$.

*This leaves the obvious question: does $\mathcal T$ consist just of the limits of polynomial functors?
A: As discussed in the comments, I'm writing here the proof of the following fact:

Let $\mathscr{X}_∞$ be the ∞-topos of sheaves on $\mathrm{FinTop}^{op}$ under the atomic topology (the topology where a sieve is covering iff it is nonempty). Then for any other ∞-topos $\mathscr{X}$, there is a natural equivalence between the ∞-category of geometric morphisms $\mathscr{X}\to \mathscr{X}_∞$ and the ∞-category of ∞-connective objects in $\mathscr{X}$.

To do so, we will need a lemma.
Lemma: Let $\mathscr{X}$ be an ∞-topos and let $X\in\mathscr{X}$ be an object. Then all the following statements are equivalent


*

*The object $X$ is ∞-connected. That is $\pi_nX\to X$ is an equivalence for each $n\ge 0$ and $X\to *$ is an effective epimorphism.

*For every $n\ge -1$ the diagonal map $X\to X^{S^n}$ is an effective epimorphism (where $S^{-1}=\varnothing$)

*For every finite space $T$ the diagonal map $X\to X^T$ is an effective epimorphism.

*For every nonempty collection of maps of finite spaces $\{T\to T'_i\}_{i\in I}$ the map $\coprod_{i\in I} X^{T_i'}\to X^T$ is an effective epimorphism.


The lemma immediately implies the fact we were looking for thanks to proposition 6.2.3.20 in Higher Topos Theory. In fact our condition 4 is exactly the condition that the left exact functor $\mathrm{FinTop}^{op}\to \mathscr{X}$
$$T\mapsto X^T$$
sends covering families to effective epimorphisms.
Proof: $4 \Rightarrow 3\Rightarrow 2$ is obvious. Let us prove $1\Leftrightarrow 2$. We know that $\pi_nX$ is the 0-truncation of $X^{S^n}\to X$ given by the evaluation at the basepoint. So $\pi_nX\to X$ is an equivalence iff $X^{S^n}\to X$ is a 0-connected. But by HTT.6.5.1.20, this is true iff the diagonal map $X\to X^{S^n}$ (which is a section of the evaluation) is -1-connected, i.e an effective epimorphism. 
In the following we will often use HTT.6.2.3.12, that is if $gf$ is an effective epimorphism, so is $g$.
$2\Rightarrow 3$ This follows from an induction on the number of cells of $T$. Let us assume that it is true for $T$ and we will show it is true for $T'=T\amalg_{S^{n-1}}D^n$. There is a cofiber sequence
$$T\to T'\to T'/T=S^n$$
and so we obtain a pullback square
$$\require{AMScd}
\begin{CD} 
X^{S^n} @>>> X^{T'}\\
 @VVV  @VVV\\
 X @>>> X^T
\end{CD}
$$
Since effective epimorphisms are stable under pullbacks (HTT.6.2.3.15), it follows that $X^{S^n}\to X^{T'}$ is an effective epimorphism. Finally, since $X\to X^{S^n}$ is an effective epimorphism by hypothesis and by HTT.6.2.3.12 effective epimorphisms are stable under composition, we are concluded.
$3 \Rightarrow 4$ This follows from HTT.6.2.3.12 applied to the compositions
$$ X\to X^{T_j'}\to \amalg_{i\in I}X^{T_i'}\to X^T$$
for some $j\in I$. $\square$
