I have a hand-wavy answer for this. These are not rigorous results.
Takeaway: The ferromagnetic XX chain is in the superfluid phase of the 1d Bose-Hubbard model (BHM). Particle number in the BHM is analogous to $S^z$ in the XX chain. This is a phase with quasi-long range order in the $S^-$ order parameter, a gapless spectrum with $\Delta \sim L^{-1}$, and power law correlations.
Bose-Hubbard model
This is an interacting model of bosons
$$
H = \sum_i\left(-t (b^\dagger_i b_{i+1} + b^\dagger_{i+1}b_i) - \mu b^\dagger_i b_i + \frac{U}{2} b^\dagger_ib^\dagger_i b_i b_i \right)
$$
Here is the intuition for the Bose-Hubbard model phase diagram (figure 3 in the paper). We start at certain values of $\mu/U$ and look at the behavior while increasing $t/U$. If $\mu /U$ is at a value such that having $N$ or $N+1$ particles at the same site have equal energy (for some $N$), then an arbitrarily small amount of hopping would drive a transition into a superfluid phase, because the ground state would prefer to have superpositions of $N$ and $N+1$ bosons on each site to gain hopping energy. This is a Bose-condensed phase which spontaneously breaks the particle number conservation symmetry.
For generic $\mu/U$ which prefers to have $N$ particles at a site, you need some finite hopping strength before the bosons condense, since there is a gap to having $N\pm 1$ particles at a site. For more discussion on BHM, refer these notes. Note that what I have described is a mean-field picture of what happens, there are some important details which are different in 1 dimension.
Spin-$1/2$ XX chain
It is straightforward to map the spin-1/2 XX chain to a BHM. A spin-$1/2$ degree of freedom is equivalent to a hard-core boson (a bosonic site with $U \to \infty$ such that it can only hold up to one particle). Therefore, the ferromagnetic XX chain becomes a BHM with $U \to \infty$, $\mu = 0$ and $t = J$. Note that this satisfies the condition that $N$ and $N+1$ bosons at one site have the same energy when $N=0$. Therefore, an arbitrarily small hopping drives a transition into a gapless superfluid phase with $\Delta \sim L^{-1}$, consistent with the exact solution from the Jordan-Wigner transformation.
Spin-$S$ XX chain
How do we generalize this to spin $S$? There is no choice of $U$ for the BHM that recovers the spin-$S$ XX chain. But, intuitively, it is believable that the idea is the same. At one site, the XX chain does not care if $S^z = -S, -S+1, \dots, S-1,S$. So an arbitrarily small amount of "hopping"
$$
S^x_{i}S^x_{i+1} + S^y_{i}S^y_{i+1} = S^+_{i}S^-_{i+1} + S^-_{i}S^+_{i+1}
$$
makes the particle number want to fluctuate and drives a transition into a superfluid.
Can we make this more believable? The best I could do is to map this to a problem of $2 S$ spin-$1/2$ degrees of freedom per site, with the mapping of operators to Pauli matrices
$$
S_i^{\mu} = \frac{1}{2}\sum_{\alpha=1}^{2S} \sigma_{i,\alpha}^\mu
$$
This satisfies all the spin-$S$ commutation relations, so is a valid representation of the original problem (note that this is a redundant description as we are introducing new degrees of freedom -- my intuition is that this is okay for the low-energy physics of the ferromagnetic case, but maybe not for the anti-ferromagnetic case).
In this language, the Hamiltonian becomes
$$
H = \frac{-J}{4}\sum_{i=1}^L \sum_{\alpha,\beta=1}^{2S} \left( \sigma^x_{i,\alpha} \sigma^x_{i+1,\beta} + \sigma^y_{i,\alpha} \sigma^y_{i+1,\beta} \right)
$$
which can be reduced (following the discussion for the spin-$1/2$ case) to a Bose-Hubbard model with $U \to \infty$, albeit with a non-standard geometry but that is not consequential for any of our arguments. This is further proof that the spin-$S$ XX chain is in the superfluid phase of the Bose-Hubbard model.