The answer is "for lots of reasons". Let me explain a couple:
1.
As several people have already noted in answers and comments, sometimes profinite groups arise naturally. For example, if $k$ is a field and $k^s$ is its separable closure, it is natural to consider the group of automorphisms of $k^s$ over $k$. One then equips this group with a topology (the weak topology, when $k^s$ is equipped with its discrete topology --- i.e. two elmenents are close it their action coincides on a large finite set of elements of $k^s$), and then discovers that this makes the automorphism group into a profinite group.
Now one can bring the pro-structure to the fore by regarding $k^s$ not just as a field extension of $k$, but as an ind-finite extension of $k$, by writing it as the inductive limit of its finite subextensions. But, while this is technically useful in some contexts (for example, in the proof the automorphism group is profinite), it is not always convenient --- there are often advantages to having $k^s$ available as a naked field, without having to bother with its ind-structure.
2.
The concept of topology is incredibly, amazingly flexible, much more so than the concept of pro-object. There are lots of illustrations of this, but one very convincing one is the theory of the adeles. Here one takes the topological product of a profinite ring with copies of $\mathbb R$ and $\mathbb C$. One obtains a locally compact ring, equips it with a Haar measure, and proceeds to do harmonic analysis. Trying to carry all this out in the language of pro-systems (say, of pro-Lie groups) would be incredibly convoluted. Indeed, in the early days of class field theory, before the introduction of the adelic view-point, this is essentially what people did: they worked explicitly with the pro-systems underlying the adeles (without using that language, of course). The introduction of ideles and adeles swept away the inherent (conceptual and notational) complexities of that view-point, and so was (and is) rightly regarded as a major advance.
It should be noted that already in $\mathbf{Set}$, the free functor $\mathbf Z^{(-)} \colon \mathbf{Set} \to \mathbf{Ab}$ does not preserve cofiltered limits. For a cofiltered diagram $D \colon \mathcal I \to \mathbf{Set}$, write $S_i$ for its value at $i \in \mathcal I$, write $S$ for its limit, and write $\pi_i \colon S \to S_i$ for the canonical projection.
There is always a map
\begin{align*}
\phi \colon \mathbf Z^{(S)} &\to \lim_\leftarrow \mathbf Z^{(S_i)}\\
\sum_{k=1}^n n_k s_k &\mapsto \left( \sum_{k=1}^n n_k\pi_i(s_k) \right)_i.
\end{align*}
Denote its $i^{\operatorname{th}}$ component by $\phi_i$. For a set $X$ and an element $z=\sum_{k=1}^n n_k x_k \in \mathbf Z^{(X)}$ with $x_k \neq x_{k'}$ for $k \neq k'$, write $\operatorname{Supp}(z)$ for $\{x_1,\ldots,x_n\}$, and denote by $|z|$ its cardinality. If $f \colon X \to Y$ is a map, we denote the induced map $\mathbf Z^{(X)} \to \mathbf Z^{(Y)}$ by $f$ as well (by abuse of notation). For $z \in \mathbf Z^{(X)}$, we have $\operatorname{Supp}(f(z)) \subseteq f(\operatorname{Supp}(z))$, so $|f(z)| \leq |z|$ with equality if and only if $f_* \colon \operatorname{Supp}(z) \to \operatorname{Supp}(f(z))$ is a bijection.
Lemma. The map $\phi$ is injective, and its image consists of those $(x_i)_{i \in \mathcal I}$ for which there exists $n \in \mathbf Z$ with $|x_i| \leq n$ for all $i \in \mathcal I$.
In other words, the image consists of the sequences $(x_i)_i$ of bounded support.
Proof. For injectivity, if $x = \sum_{k=1}^n n_ks_k \in \ker(\phi)$ is such that $s_k \neq s_{k'}$ for $k \neq k'$, then there exists $i \in \mathcal I$ such that $\pi_i(s_k) \neq \pi_i(s_{k'})$ for $k \neq k'$ (here we use that the sum is finite and that $\mathcal I$ is cofiltered). Then $\phi_i(x) = \sum_{k=1}^n n_k\pi_i(s_k)$ is zero by assumption, so all $n_k$ are zero.
For the image, it is clear that $|\phi_i(x)| \leq n$ for all $i \in \mathcal I$ if $x \in \mathbf Z^{(S)}$ has $|x| = n$. Conversely, if $|x_i| \leq n$ for all $i \in \mathcal I$, then decreasing $n$ if necessary, we may assume $|x_{i_0}| = n$ for some $i_0 \in \mathcal I$. Then $|x_i| = n$ for all $f \colon i \to i_0$ since $n = |x_{i_0}| = |D(f)(x_i)| \leq |x_i| \leq n$. Replacing $\mathcal I$ by the coinitial segment $\mathcal I/i_0$ we may therefore assume $|x_i| = n$ for all $i \in \mathcal I$.
Constancy of $|x_i|$ means that for every morphism $f \colon i \to j$ in $\mathcal I$, the map $f_* \colon \operatorname{Supp}(x_i) \to \operatorname{Supp}(x_j)$ is a bijection. Setting $T = \lim\limits_\leftarrow \operatorname{Supp}(x_i)$, we see that each projection $\pi_i \colon T \to \operatorname{Supp}(x_i)$ is a bijection. Functoriality of the limit gives an injection $T \hookrightarrow S$, giving elements $s_1,\ldots,s_n \in S$ such that $\pi_i(\{s_1,\ldots,s_n\}) = \operatorname{Supp}(x_i)$ for all $i \in \mathcal I$. The coefficients must also be constant under the bijections $\operatorname{Supp}(x_i) \to \operatorname{Supp}(x_j)$, so we get an element $x = \sum_{k=1}^n n_ks_k \in \mathbf Z^{(S)}$ with $\phi(x) = (x_i)_i$. $\square$
Example. An example where $\phi$ is not surjective: let $S = \mathbf Z_3$ with $S_i = \mathbf Z/3^i$. Define the element $(x_i)_i \in \prod_i \mathbf Z^{(S_i)}$ where $x_i \in \mathbf Z^{(S_i)} = \mathbf Z^{S_i}$ has coordinates (for $k \in \{0,\ldots,3^i-1\}$) given by
$$x_{i,k} = \begin{cases} 1, & \text{the first $3$-adic digit of } k \text{ is } 1, \\ -1, & \text{the first $3$-adic digit of } k \text{ is } 2, \\ 0, & k=0.\end{cases}$$
These form an element of $\lim\limits_\leftarrow \mathbf Z^{(S_i)}$: any fibre of $\mathbf Z/3^{i+1} \to \mathbf Z/3^i$ above $k \in \{0,\ldots,3^i-1\}$ consists of $\{k,3^i+k,2 \cdot 3^i+k\}$, of which $x_{i+1,k} = x_{i,k}$ and the others are $1$ and $-1$ since $3^i+k$ starts on $1$ and $2 \cdot 3^i+k$ starts on $2$.
Since $\operatorname{Supp}(x_i) = S_i \setminus \{0\}$, we see that $(x_i)_i$ does not have bounded support, hence is not in the image of $\phi$.
Corollary. For any presheaf topos $\mathbf T = \mathbf{PSh}(\mathscr C) = [\mathscr C^{\operatorname{op}},\mathbf{Set}]$ on a small nonempty category $\mathscr C$, the free functor $\mathbf Z^{(-)} \colon \mathbf T \to \mathbf{Ab}(\mathbf T)$ does not preserve cofiltered limits.
Proof. Let $f \colon \mathscr C \to *$ be the map to the terminal category $*$, on which $\mathbf{PSh}(*) = \mathbf{Set}$. Note that $f$ has a section $g$ since $\mathscr C$ is nonempty. We saw above that in $\mathbf{Set}$, the free functor $\mathbf Z^{(-)}$ does not preserve cofiltered limits. The pullback $f^* \colon \mathbf{Set} \to \mathbf{PSh}(\mathscr C)$ takes a set $S$ to the constant sheaf $\underline{S}$ on $\mathscr C$. Since limits, colimits, and free abelian group objects in presheaf categories are pointwise, $f^*$ commutes with formation of limits, colimits, and free abelian group objects. Thus pulling back everything along $f^*$ gives the natural map $\underline{\mathbf Z}^{(\underline S)} \to \lim\limits_\leftarrow \underline{\mathbf Z}^{(\underline S_i)}$, which is not an isomorphism since it isn't after applying the section $g^*$ (this is just evaluation at an object $g(*) \in \mathscr C$). $\square$
The class of topoi where the free functor does not commute with cofiltered limits is probably much larger still. For instance, my example does not include condensed sets, which is close to a presheaf topos but not quite. I'm not even sure what a topos would look like where these do commute!
Best Answer
I'm sorry for being cryptic.
The subtle point in the construction is that the maps $T_n\to T_{n,j}$ are not all surjective, i.e. one cannot construct this pro-system as a system of quotients.
By induction on $n$, one constructs a cofiltered system of $n$-skeletal simplicial hypercovers $T_{\bullet,j}^{(n)}\to S_j$ (of finite sets by finite sts) whose limit is the $n$-skeleton of $T_\bullet\to S$, where each inductive step is refining the previous one; in the end, one takes the limit over $n$ of the induced coskeletal simplicial hypercovers.
In the first step, one needs to write the surjection $T_0\to S$ as a limit of surjections of finite sets; this can be done by simply taking the cofiltered system of compatible finite quotients. Next, one needs to write $T_1$ as a cofiltered limit of surjections $T_{1,j}\to T_{0,j}\times_{S_j} T_{0,j}$. For this, we prove
Admitting the lemma for the moment, one can find the desired $T_{1,j}\to T_{0,j}\times_{S_0} T_{0,j}$. But for higher $n$, one is in the similar situation, for the surjection $T_n\to (\mathrm{cosk}_{n-1} \mathrm{sk}_{n-1} T_\bullet)_n$ where the target already has a given presentation as a cofiltered limit of finite sets.
Thus, it remains to prove the lemma. For this, one considers the category $J$ whose objects $j\in J$ consist of an object $i\in I$ and a factorization of $Y\to X\to X_i$ into $Y\to Y_j\to X_i$ where $Y_j$ is finite and $Y_j\to X_i$ is surjective. A morphism $j\to j'$ is given by a map $i\to i'$ and a map $Y_j\to Y_{j'}$ making the obvious diagrams (from $Y$, and to $X_i$) commute.
One needs to see that $J$ is cofiltered, and that $Y$ is the limit of the $Y_j$'s. Given two $j,j'\in J$, one can first arrange via cofilteredness of $I$ that they map to the same $i\in I$. Then we have two maps $Y\to Y_j\to X_i$ and $Y\to Y_{j'}\to X_i$ where $Y_j\to X_i$ and $Y_{j'}\to X_i$ are surjective. We get a similar factorization over $Y_j\times_{X_i} Y_{j'}$, which is still finite and surjective over $X_i$, so we find some $j''$ dominating $j$ and $j'$. Now, given two morphisms $f,g: j\to j'$, we need to find some $j''$ mapping to $j$ equalizing $f$ and $g$. Again, we can assume that the diagram $f,g: j\to j'$ gets contracted to $i\in I$ (using that $I$ is cofiltered). Then we get two maps $Y_j\to Y_{j'}$ over $X_i$, and $Y$ factors over the equalizer $Z\subset Y_j$. This is finite, but not necessarily surjective over $X_i$! Still, as $Y\to X$ is surjective, there is some $i'\to i$ such that $Z\times_{X_i} X_{i'}\to X_{i'}$ is surjective. This defines the desired object $j''\in J$.
It remains to see that $Y\to \mathrm{lim}_j Y_j$ is bijective. Injectivity is easy to see (for any finite quotient $Y'$ of $Y$, $Y\to Y'\times X_i\to X_i$ defines an object of $J$). For surjectivity, the argument at the end of the last paragraph applies (one needs to see that for any $j$ there is some $j'$ such that $Y_{j'}\to Y_j$ factors over the image of $Y\to Y_j$, which was proved there).