This is not really an answer to the question, but I thought it might be helpful to restate Hikita's main result (at least for $q=1$) in a way that may make it easier for people to quickly understand what it is saying, and then write their own code to verify it computationally for small examples.
As a preliminary step, given positive real parameters $a_1,b_1,a_2,b_2,\ldots,a_m,b_m$, let us describe a random variable $X(a_1,b_1,\ldots,a_m,b_m)$ whose value lies in the set $\{1,2,\ldots,m+1\}$. Arrange $2m+1$ real numbers
$$ A_0 < B_1 < A_1 < B_2 < A_2 < B_2 < \cdots < A_m < B_m $$
on the real line, such that for all $i$, $a_i = B_i - A_{i-1}$ and $b_i = A_i - B_i$. We throw $m$ darts at certain subintervals of the real line. The first dart is thrown at $[A_0,A_1]$, meaning that we pick a uniformly random number in the interval $[A_0,A_1]$. The parameters for the second dart throw depend on the outcome of the first dart throw. If the first dart lands on the Left (or $L$ for short), meaning in $[A_0,B_1)$, then we Leave the left endpoint alone, and throw the second dart at the interval $[A_0,A_2]$. If the first dart lands on the Right (or $R$ for short), meaning in $[B_1,A_1]$, then we Ratchet up the left endpoint, and throw the second dart at the interval $[A_1,A_2]$. In general, the $i$th dart is thrown at an interval whose right-hand endpoint is $A_i$ and whose left-hand endpoint is either the same as it was for the previous throw (if the outcome of the previous throw was $L$) or is $A_{i-1}$ (if the outcome of the previous throw was $R$). If the $i$th dart lands in $[B_i,A_i]$ then we declare that the $i$th outcome is $R$; otherwise we declare that the $i$th outcome is $L$. In this way we obtain a sequence of $m$ letters, each of which is either $L$ or $R$. Finally, we let $X(a_1,b_1,\ldots,a_m,b_m)$ be 1 plus the index of the rightmost $R$ (if there is no such index then we let $X(a_1,b_1,\ldots,a_m,b_m) = 1$).
As an example, the reader can check that for $m=2$,
$$\eqalign{ \Pr(X=1) &= \biggl( {a_1\over a_1+b_1}\biggr) \biggl({a_1+b_1+a_2\over a_1 + b_1 + a_2 + b_2}\biggr)\cr
\Pr(X=2) &= \biggl({b_1\over a_1+b_1}\biggr) \biggl({a_2\over a_2+b_2}\biggr) \cr
\Pr(X=3) &= \biggl({b_2\over a_2+b_2}\biggr) \biggl({a_2 + b_1+b_2\over a_1+a_2+b_1 + b_2}\biggr)\cr
}$$
The formula for $\Pr(X=3)$ is not so obvious, but it is correct; this is the content of Hikita's Lemma 6.
[EDIT: In the important special case where the $a_i$ and $b_i$
are all positive integers, my colleague Doug Jungreis came up with an
equivalent but simpler definition of $X(a_1,b_1,\ldots,a_m,b_m)$.
For all $i$, let $s_i := \sum_{j=1}^i (a_j + b_j)$, and let
$M := s_m$. Generate a uniformly random permutation $\pi \in \mathfrak{S}_M$.
Then let $X$ be the smallest $i\ge 1$ such that the largest element of
$\{\pi_{s_{i-1}+1}, \pi_{s_{i}+2}, \ldots, \pi_{s_j}\}$ lies in the subset
$\{\pi_{s_{i-1}+1}, \pi_{s_{i}+2}, \ldots, \pi_{s_{j-1}+a_j}\}$ for all $j\ge i$.
If there is no such $i$, then let $X = m+1$. The proof that this
description is equivalent is not too difficult; the main subtle point
to check is that generating a single random permutation can simulate
all $m$ dart throws, even though we are "reusing randomness" when
intervals overlap.]
Now suppose we are given a (naturally labeled) unit interval graph $G$ on the vertex set $\{1,2,\ldots,n\}$. For $i=1,2,\ldots,n$, let $h'_i$ be the number of vertices less than $i$ that are not adjacent to $i$. (Hikita uses the notation $e_i$ rather than $h'_i$, but I want to reserve the notation $e_i$ for the $i$th elementary symmetric function.) For example, if $G$ is the path on four vertices, with edges $\{1,2\}$ and $\{2,3\}$ and $\{3,4\}$, then $h'_1=0$, $h'_2 = 0$, $h'_3 = 1$, and $h'_4 = 2$. We now describe a random process (depending on $G$) for generating a standard Young tableau with $n$ boxes, one box at a time.
There is no choice about where to put box 1. When it comes time to place box $i$ (for $i>1$), we define (for all $j\ge 1$) $\delta_j$ to be 1 if the partial tableau constructed so far has a box in column $j$ whose entry is greater than $h'_i$. Otherwise, we let $\delta_j = 0$. The sequence $(\delta_j)$ consists of alternating segments of 0's and 1's, so we can write it uniquely in "compressed form" as $1^{b_0}0^{a_1}1^{b_1}0^{a_2}1^{b_2} \cdots 0^{a_m}1^{b_m}0^\infty$ for some nonnegative integer $m$ and some nonnegative integers $b_0, \ldots, b_m$ and $a_1,\ldots, a_m$ (where $b_0$ may be zero but all the rest are strictly positive). Now we generate a random number $X(a_1,b_1,\ldots,a_m,b_m)$ as described earlier (if $m=0$ then we set $X=1$). Finally, we put the box with entry $i$ in the column corresponding to the first $0$ of $0^{a_X}$ (if $X=m+1$ then we interpret $0^{a_X}$ as the $0^\infty$ at the end).
Continuing our example of the path with four vertices, at step 2, the delta sequence is $1^10^\infty$, so $m=0$ and we have no choice but to put box 2 at the start of $0^\infty$, i.e., in column 2 of the standard Young tableau we are building. At step 3, the delta sequence is $1^00^11^10^\infty$, so $m=1$ and with probability 1/2 we put box 3 in column 1 (call this Case A) and with probability 1/2 we put box 3 in column 3 (call this Case B). At step 4, if we are in Case A, then the delta sequence is $1^10^\infty$ and we are forced to put box 4 in column 2; if we are in Case B then the delta sequence is $1^00^21^10^\infty$, so we put box 4 in column 1 with probability 2/3 and in column 4 with probability 1/3. All told, three of the SYT with four boxes have a positive probability of being generated, and their probabilities are 1/2, 1/3, and 1/6 respectively.
Let $p_T(G)$ denote the probability that SYT $T$ is generated. Then Hikita's main theorem (for $q=1$) is that for all partitions $\lambda = (\lambda_1, \lambda_2, \ldots)$,
$${c_\lambda\over \prod_i \lambda_i!} = \sum_{T\colon {\text{shape}}(T)=\lambda} p_T(G),$$
where $c_\lambda$ is the coefficient of $e_\lambda$ in the elementary symmetric function expansion of the chromatic symmetric function $X_G$.
