That's a good question! I think Barwick and Haine have thought much more about this, and maybe they already know the answer? What I say below is definitely known to them. Also beware that I've written the below in a stream of consciousness, not quite knowing where it will go when I started.
I'll write "anima" for what is variously called homotopy types/spaces/$\infty$-groupoids/..., and denote their $\infty$-category $\mathrm{An}$($=\mathcal S$). We can also consider the $\infty$-category $\mathrm{CondAn}=\mathrm{Cond}(\mathrm{An})$ of condensed anima (this is, by the way, also the animation of the category of condensed sets). If $X\in \mathrm{CondAn}$ is a condensed anima, then $\pi_0 X$ is a condensed set, and for any point $x\in X$, one can define homotopy groups $\pi_i(X,x)$ for $i\geq 1$, which are condensed groups (abelian for $i\geq 2$). Slightly more generally, if $S$ is any profinite set and $g: S\to X$ is any map, one can define a group object $\pi_i(X,g)\to S$ in condensed sets over $S$, whose fibre over any $s\in S$ is $\pi_i(X,g(s))$. Then a map of condensed anima is an equivalence if and only if it induces an equivalence on $\pi_0$ and all $\pi_i$ for $i\geq 1$ (at all base points, including profinite families of basepoints).
So, just like in a very very crude approximation an anima $X$ is something like the collection $\pi_0 X,\pi_1 X,\pi_2 X,\ldots$ of a set, a group, and abelian groups, a condensed anima is something like a collection of a condensed set, a condensed group, and condensed abelian groups. In particular, already $\pi_0 X$ can be an interesting topological space like a manifold, so a space. This is why we do not say "condensed space", as then it would seem like forgetting to condensed sets should forget the "space" structure, but it rather forgets "abstract homotopy" structure.
Now the following seems like the obvious "$\infty$-categorical compact Hausdorff spaces":
Definition. A condensed anima $X$ is "compact Hausdorff" if $\pi_0 X$ and all $\pi_i X$ for $i\geq 1$ are compact Hausdorff.
Recall here that compact Hausdorff spaces embed fully faithfully into condensed sets. The second statement means more precisely that for all profinite sets $S$ with a map $g: S\to X$, the group object $\pi_i(X,g)\to S$ in condensed sets over $S$ is compact Hausdorff. (This is a little stronger than only asking it at all fibres.)
So in this case $\pi_0 X$ is a compact Hausdorff space, $\pi_1 X$ is a compact Hausdorff group, and $\pi_2 X,...$ are compact Hausdorff abelian groups.
It turns out that there is a nice characterization of "compact Hausdorff" condensed anima. In fact, there is a general topos-theoretic notion of "coherent"="qcqs" objects. This is usually studied for $1$-topoi, but it generalizes easily to $n$-topoi. Basically, an object is quasicompact if any cover admits a finite subcover; it is quasiseparated if the diagonal is quasicompact; it is 2-quasiseparated if the diagonal is quasiseparated; etc.; and coherent = quasicompact and $n$-quasiseparated for all $n\geq 1$. Then coherent condensed sets are exactly compact Hausdorff spaces, and:
Proposition. Coherent condensed anima are exactly the "compact Hausdorff" condensed anima.
Note: In a $1$-topos, coherent objects often agree with the finitely presented objects, but this fails dramatically for $\infty$-topoi, where coherence and finite presentation are two quite different finiteness conditions. In the case of anima, coherence means finite homotopy groups, while finite presentation should mean generated under finite colimits from the point; these are very different notions. As already discussed in the comments, the "finite homotopy groups" condition seems more relevant for the question.
Now we have a good notion of "$\infty$-categorical compact Hausdorff spaces". The question however started from a different angle, namely as trying to describe it via a monad on anima. The good news is:
Proposition. Compact Hausdorff condensed anima are monadic over anima.
This can be deduced from Barr-Beck-Lurie, although it takes some work.
It remains to understand the monad (and see whether it can be described as a codensity monad). The monad takes an anima $X$ to $\lim_{X\to Y} Y$ where the diagram is over all maps from $X$ to a compact Hausdorff condensed anima $Y$: This computes the desired left adjoint. Assume for the moment that the diagram category was small; then this limit is still a compact Hausdorff condensed anima: The compact Hausdorff condensed anima are stable under all small limits, as they are stable under finite limits and all small products. Now the diagram category is not actually small, so one has to argue slightly more carefully to see the existence of the left adjoint.
If $X$ is actually a set, then one can show that the left adjoint is still the same as usual, given by the Stone-Čech compactification. This is the same as $\lim_{X\to Y} Y$ where we restrict $Y$ to be a finite set. Ultimately, the possibility to restrict $Y$ to finite sets here -- coming from the fact that the Stone-Čech compactification is totally disconnected, and totally disconnected compact Hausdorff spaces are pro-finite -- is what makes it possible to describe compact Hausdorff spaces in terms of the codensity monad for $\mathrm{FinSet}\hookrightarrow \mathrm{Set}$.
The first interesting new case is $X=K(G,1)$, for some discrete group $G$. Ignoring higher homotopy groups, we are then interested in the universal compact group $H$ with a map $G\to H$. In general, this is known as the "Bohr compactification" of $G$. If $G=\mathbb Z$, then we look for the free compact group on one generator. This is necessarily abelian, and then one can use Pontrjagin duality to actually determine this (I hope I didn't screw this up): Take $\prod_{\mathbb R/\mathbb Z}\mathbb R/\mathbb Z$, the product of $\mathbb R/\mathbb Z$ (as a discrete set) many copies of the circle $\mathbb R/\mathbb Z$, with its tautological "diagonal" element, and take the closed subgroup generated by this element.
What we see from the example is that already for the anima $X=K(\mathbb Z,1)$ (aka the circle), the monad takes an extremely complicated value (note that we were ignoring higher homotopy groups, but the computation of $\pi_1$ is correct), that in particular is not itself totally disconnected, and so cannot be written as a limit of finite anima. So I gather that these "$\infty$-categorical compact Hausdorff spaces" cannot be described in the way the question started.
This, then again, begs the question what algebras for the monad in the question are!
Well, I don't know the precise answer, but one can also consider "totally disconnected compact Hausdorff" condensed anima, asking now that all $\pi_i X$ are totaly disconnected compact Hausdorff. So $\pi_0 X$ is a profinite set, $\pi_1 X$ is a profinite group, and $\pi_2 X,\ldots$ are profinite abelian groups.
Proposition. "Totally disconnected compact Hausdorff condensed $n$-truncated anima" are equivalent to the Pro-category of $n$-truncated anima with finite homotopy groups.
One can also pass to the limit $n\to \infty$ in some sense, but has to be careful as this does not exactly commute with passage to Pro-categories. It is still true that any totally disconnected compact Hausdorff condensed anima $X$ maps isomorphically to the $\lim_{X\to Y} Y$ where $Y$ runs over anima with finite homotopy groups.
Now totally disconnected compact Hausdorff condensed anima are not monadic anymore over anima, but the forgetful functor still detects isomorphisms, and has a left adjoint, so gives rise to a monad on anima, and totally disconnected compact Hausdorff condensed anima embed fully faithfully into algebras over this monad. And this monad, by the last paragraph, can be identified with the codensity monad for the inclusion $\mathrm{An}^{\mathrm{coh}}\hookrightarrow \mathrm{An}$ of coherent anima (=anima with finite homotopy groups) into all anima.
So, if I'm not screwing this up, then the category of algebras over this monad is some kind of hull of totally disconnected compact Hausdorff condensed anima (including all geometric realizations that are split on underlying anima); this hull is contained in compact Hausdorff condensed anima.
In summary, if one takes "finite anima" in the question to mean "finite homotopy groups", then this gives rise to a monad whose algebras lie somewhere between totally disconnected compact Hausdorff condensed anima, and all compact Hausdorff condensed anima. I think they definitely include all those for which $\pi_0 X$ is arbitrary compact Hausdorff, but $\pi_i X$ for $i\geq 1$ is totally disconnected.
Hmm... OK, let me make the following:
Conjecture: Algebras over the codensity monad for $\mathrm{An}^{\mathrm{coh}}\hookrightarrow \mathrm{An}$ are exactly those compact Hausdorff condensed anima $X$ for which all $\pi_i X$ for $i\geq 1$ are totally disconnected.
I'm willing to conjecture this for the following reason: while one can obtain all compact Hausdorff spaces as quotients of profinite sets by closed equivalence relations, nothing like this happens for groups: a quotient of a profinite group by a closed equivalence relation is still a profinite group.
I think I can give a concrete description of these maps, and have an approach to proving faithfulness.
Let $V^{*n}$ be the $n$'th dual of $V$.
A map $V^{*n} \to V^{*m}$ is, by definition, the same thing as a bilinear form $V^{*n} \times V^{*(m-1)} \to k$.
There are two obvious ways to produce such a bilinear form:
from a map $V^{*n} \to V^{* (m-2)}$, i.e. from a bilinear form $V^{*n} \times V^{* (m-3)} \to k$,
from a map $V^{* (m-1)} \to V^{*(n-1)}$, i.e. from a bilinear form $V^{*(m-1)} \times V^{* (n-2)} \to k$.
Note that each of these methods reduces one of the "exponents" by two. If we iteratively apply these methods, we can stop at the obvious bilinear form $V^{*} \times V \to k$ or $V \times V^* \to k$. Which one we stop at will depend on which of $n$ or $m-1$ is odd (exactly one must be odd and one even for a natural map to exist, even in the finite-dimensional case).
So we apply method one $\lfloor \frac{m-1}{2} \rfloor$ times and method two $\lfloor \frac{n}{2} \rfloor$ times. The total number of maps produced this way is $$\binom{ \lfloor \frac{n}{2} \rfloor + \lfloor \frac{m-1}{2} \rfloor}{ \lfloor \frac{m-1}{2} \rfloor},$$ and this set of orders is naturally in bijection with the order-preserving maps appearing in the walking adjunction in the description Martin Brandenburg linked in the comments. Thus I think these maps are the ones produced by the walking adjunction.
To check that these maps are linearly independent, it suffices to prove that the linear span of the canonical maps produced by method 1 does not intersect the linear span of the canonical maps produced by method 2. Then these two linear spaces would be independent, and we would break them further into independent summands, and so on, until finally breaking them into independent one-dimensional spaces.
To prove this, we can try the following lemma:
Let $W$ and $U$ be vector spaces over $k$. A bilinear form $W^* \times U^* \to k$ that arises from a map $W^* \to U$ and arises from a map $U^* \to W$ in fact arises from an element of $U \otimes W$.
Let $B$ be such a bilinear form satisfying $B(w,u)= u(a(w))=w(b(u))$ for suitable maps $a: W^\vee \to U$ and $b: U^\vee \to W$. Fix bases $\{e_i \}_{i \in I}$ of $W$ and $\{f_j\}_{j \in J}$ of $U$. Let $e_i^\vee$ and $f_j^\vee$ be the dual elements of $W^\vee$ and $U^\vee$, not necessarily forming a dual basis. Form a graph with vertices $I \cup J$ where $i$ and $j$ are connected by an edge if $B( e_i^\vee, f_j^\vee) \neq 0$. Then each vertex has finite degree since $i$ is only connected to indices of basis vectors summing to $a( e_i^\vee)$ and $j$ is only connected to indices of basis vectors summing to $b(f_j^\vee)$.
So either there are finitely many edges or there exists an infinite sequence of edges where there vertices of distinct edges in the sequence are not adjacent.
In the first case, there is an element $x\in V \otimes W$ whose value on $e_i^\vee \times e_j^\vee$ is equal to $B(e_i^\vee, e_j^\vee)$ for all $i,j$, hence by the continuity of $B$ in $W^\vee$, the value of $x$ on $w \times e_j^\vee$ is equal to $B(w, e_j^\vee)$ for all $w, j$ and thus by the continuity of $B$ in $U^\vee$, the value of $x$ on $w \times u$ is equal to $B(w,u)$ for all $w,u$, and we are done.
In the second case, let $S$ be set of vertices of the edges in that sequence that lie in $W$. Let $w$ be the linear form on $W$ that sends $e_i$ to $1$ if $i$ is in $S$ and $0$ otherwise. Then $B(w, e_j^\vee) = w ( b(e_j^\vee))$ is the sum of all the entries of $w(e_j^\vee)$ with indices in $S$. For $j$ any vertex of the edge in the sequence lying in $U$, exactly one nonzero entry of $w(e_j^\vee)$ lies in $S$ so $B(w, e_j^\vee) \neq 0$ and thus $B(w,e_j^\vee)$ is nonzero for infinitely many $j$, contradicting $B( w,e_j^\vee) =e_j^\vee ( a(w))$. QED
In addition, we need the following plausible claim:
There are no nonzero canonical (i.e. functorial) elements of $V^{*a} \otimes V^{*b}$ for any $a,b$.
Combining these, any canonical map in the intersection of the two subspaces of the space of bilinear forms on $V^{*n} \times V^{*(m-1)}$ must come from an element of $V^{ * (n-1) } \otimes V^{* (m-2)}$, which is unique, hence also canonical, and thus must be zero.
Best Answer
Tom, I believe $(-)^\ast: \mathbf{Vect}^{op} \to \mathbf{Vect}$ is monadic, essentially because all objects in $\mathbf{Vect}$, in particular $k$ as a module over $k$ as ground field, are injective.
For instance, to check that $(-)^\ast$ reflects isomorphisms, suppose $f: V \to W$ is any linear map. We have two short exact sequences
$$0 \to \ker(f) \to V \to im(f) \to 0$$
$$0 \to im(f) \to W \to coker(f) \to 0$$
Because $k$ is injective, the functor $(-)^\ast = \hom(-, k)$ preserves short exact sequences:
$$0 \to im(f)^\ast \to V^\ast \to \ker(f)^\ast \to 0$$
$$0 \to coker(f)^\ast \to W^\ast \to im(f)^\ast \to 0$$
and if $f^\ast$, the composite $W^\ast \to im(f)^\ast \to V^\ast$, is an isomorphism, then $W^\ast \to im(f)^\ast$ is injective, which forces $coker(f)^\ast = 0$ and therefore $coker(f) = 0$. By a similar argument, $\ker(f) = 0$. Therefore $f$ is an isomorphism.
The remaining hypotheses of Beck's theorem (in the form given in Theorem 2, page 179, of Mac Lane-Moerdijk) are similarly easy to check. Obviously $\mathbf{Vect}^{op}$ has coequalizers of reflexive pairs since $\mathbf{Vect}$ has equalizers. And $(-)^\ast: \mathbf{Vect}^{op} \to \mathbf{Vect}$ (which has a left adjoint, as pointed out) preserves coequalizers; this is equivalent to saying that $\hom(-, k)$, as a contravariant functor on $\mathbf{Vect}$, takes equalizers to coequalizers, or takes kernels to cokernels, but that's the same as saying that $k$ is injective, so we're done.
Oh, incidentally, double dualization is not a commutative or monoidal monad, if I recall correctly.
Edit: In a comment below, Tom asks for a more concrete description of $\mathbf{Vect}^{op}$ along the lines of topological algebra. I suspect the way to go is to see $\mathbf{Vect}$ as the Ind-completion (or Ind-cocompletion) of the category of finite-dimensional vector spaces, and therefore $\mathbf{Vect}^{op}$ as the Pro-completion of the opposite category, which is again $\mathbf{Vect}_{fd}$. I think I've seen before a result that this is equivalent to the category of topological $k$-modules which arise as projective limits of (cofiltered diagrams of) finite-dimensional spaces with the discrete topology, or something along similar lines, but I'd have to look this up to be sure. There might be pertinent material in Barr's Springer Lecture Notes on $\ast$-autonomous categories, but again I'm not sure.
Edit 2: Ah, found it. $\mathbf{Vect}^{op}$ is equivalent to the category of linearly compact vector spaces over $k$ and continuous linear maps. See Theorem 3.1 of this paper for example: arxiv.org/pdf/1202.3609. The result is credited to Lefschetz.