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.
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.
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 an exchangeable 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 problem, and it is NP-hard in general. This suggests that also determining $N$ may be challenging. It feels similar to hypergraph Turán problems.
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 $30$ in total. This is how the upper bound of $\frac{30}{60}$ arises.
Additional thoughts:
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: 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.Concerning the upper bound, I don't know if $N = 30$ is achievable. The largest value I've found is $N = 28$, achieved by $$\tag{2} X_1 = X_2 = X_3 = 0, \qquad X_4 = 5, \qquad X_5 = X_6 = 9$$
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.