In this answer, I will derive the following bounds:
$$0.367 \approx \frac{208}{567} \le \sup_\mu \mathbf{P}[X_1 + X_2 + X_3 < 2 X_4] \le \frac{30}{60} = \frac{1}{2}.$$

Unfortunately the upper bound is no better than in the question. But I will explain my attempts in the hope that someone else can squeeze more out of this method than I've been able to, in particular with regards to the upper bound.

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**Lower bound:** Start with $\mu = \frac{1}{2} \delta_0 + \frac{1}{6} \delta_5 + \frac{1}{3} \delta_9$. Then some tedious calculations show that
$$\mathbf{P}[X_1 + X_2 + X_3 < 2 X_4] = \frac{26}{81}, \qquad \mathbf{P}[X_1 + X_2 + X_3 = 2 X_4] = \frac{1}{8}.$$
Therefore applying the trick from jlewk's answer achieves
$$\mathbf{P}[X_1 + X_2 + X_3 < 2 X_4] = \frac{26/81}{1-1/8} = \frac{208}{567}.$$
I have no doubt that further tweaking can still improve this value.

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**Upper bound:** Let's consider an extension of the argument for the upper bound of $\frac{1}{2}$ given in the question. To this end, consider independent $X_1, \dots, X_6 \sim \mu$. Then their joint distribution is invariant under permutations, and in particular the probability
$$\mathbf{P}[X_i + X_j + X_k < 2 X_\ell]$$
is independent of the choice of the indices as long as these are all distinct. I will refer to the $X_i + X_j + X_k < 2 X_\ell$ as "versions" of the original inequality, so there are $6 \cdot \binom{5}{3} = 60$ versions.

If $N$ denotes the maximal number of versions that can be jointly satisfied, then I claim that $\frac{N}{60}$ is an upper bound on the original problem. Indeed considering whether each inequality holds defines a distribution on $\{0,1\}^{60}$, where the probability of each sequence with more than $N$ many $1$s is zero. The claim then follows by the fact that the symmetry group acts transitively on the components of $\{0,1\}^{60}$.

Thus the remaining problem is to determine the value of $N$, and this looks difficult. In fact, determining the maximal number of jointly satisfiable inequalities is known as the [maximum feasible subsystem](https://www.sciencedirect.com/science/article/pii/030439759400254G) problem, and it is NP-hard in general. This suggests that also determining $N$ may be challenging. It feels similar to [hypergraph Turán problems](https://people.maths.ox.ac.uk/keevash/papers/turan-survey.pdf).

One approach to get an upper bound on $N$ may be to assume without loss of generality that $X_1 < \dots < X_6$, and consider which quadruples $(i,j,k,\ell)$ can appear among the valid versions of the inequality. Let us assume $i < j < k$ without loss of generality and keep these fixed. Then the fact that $X_j < X_\ell$ shows that $\ell > j$ is necessary, since otherwise we get
$$X_i + X_j + X_k \ge X_i + 2 X_\ell \ge 2 X_\ell.$$
It follows that the number of possible $\ell$'s for a given triple $(i,j,k)$ is at most $5 - j$.
Enumerating thus the number of possible $\ell$ for each of the $20$ triples $(i,j,k)$ gives $30$ versions in total that can be jointly satisfied.
This is how the upper bound of $\frac{30}{60}$ arises.

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**Additional thoughts:** 

1. It's not hard to see that the proof of the upper bound is an instance of a general method for deriving upper bounds on such probabilities based on [exchangeability](https://en.wikipedia.org/wiki/Exchangeable_random_variables): for $X_1, \ldots, X_n$, consider how many versions of the inequality can be satisfied jointly. The fraction of these is then an upper bound on the desired quantity.  
  I believe that these upper bounds are tight as $n \to \infty$ thanks to the de Finetti theorem, but I haven't attempted a complete proof of this yet.

1. Concerning the upper bound, I don't know if $N = 30$ is achievable. The largest value I've found is $N = 28$, achieved by
$$\tag{2} X_1 = X_2 = X_3 = 0, \qquad X_4 = 5, \qquad X_5 = X_6 = 9$$

1. This is also how I had found the $\mu = \frac{1}{2} \delta_0 + \frac{1}{6} \delta_5 + \frac{1}{3} \delta_9$ which appears in the new lower bound: this distribution is given by the relative frequencies in (2), and my general intuition for the de Finetti theorem indicates that such distributions are good candidates.