Work in progress Uncompleted proof of a special case of the desired result.
Let us call a sequence $(a_0, \ldots, a_n)$ as unimodal with mode
$M$ if it enjoys the property that
$$a_0 < a_1 < \cdots < a_{M-1} < a_M > a_{M+1} > \cdots > a_n$$
where $1 \leq M < n$.
Let $\{Y_i\}$, denote a sequence of be n i.i.d. discrete random variables
taking on values $0,1,2$ with probabilities $p$, $q$, and $r=1−p−q$
respectively where we assume that $p < q > r$ so that $(p,q,r)$ is
a unimodal sequence with mode $1$. Define $X_n = \sum_{i=1}^n Y_i$.
Then, $X_n$ is a discrete random variable taking on value
$i$ with probability $p_n(i)$, $0 \leq i \leq 2n$.
Claim: The sequence
$(p_n(0), p_n(1), \ldots, p_n(2n))$ is unimodal.
We prove the claim by induction.
Basis: The result holds for $n = 1$ since we have assumed that
$(p_1(0), p_1(1), p_1(2)) = (p,q,r)$ is unimodal.
Induction hypothesis: $(p_n(0), p_n(1), \ldots, p_n(2n))$ satisfies
$$p_n(k) > p_n(k-1) ~\text{for}~ 1 \leq k \leq M_n, ~~
p_n(k-1) > p_n(k)~\text{for}~ M_n < k < 2n.$$
Now consider the sequence $(p_{n+1}(0), p_{n+1}(1), \ldots, p_{n+1}(2{n+2}))$.
Since
$$p_{n+1}(k) = p_n(k)\cdot p + p_n(k-1)\cdot q + p_n(k-2)\cdot r$$
(where we use $p_n(-2) = p_n(-1) = p_n(2n+1) = p_n(2n+2) = 0$), we have
that for $1 \leq k \leq M_n$,
$p_{n+1}(k) -p_{n+1}(k-1)$ is the sum of three positive terms and therefore
$(p_{n+1}(0), p_{n+1}(1), \ldots, p_{n+1}(M_n))$ is an increasing
sequence. Similarly, for $M_n +2 \leq k \leq 2n+2$,
$p_{n+1}(k) -p_{n+1}(k-1)$ is the sum of three negative terms and therefore
$(p_{n+1}(M_n+2), p_{n+1}(M_n+3), \ldots, p_{n+1}(2n+2))$ is a decreasing
sequence.
If $p_{n+1}(M_n) < p_{n+1}(M_n+1) < p_{n+1}(M_n+2)$, then
$(p_{n+1}(0), p_{n+1}(1), \ldots, p_{n+1}(2{n+2}))$ is unimodal with
mode $M_n+2$.
If $p_{n+1}(M_n) < p_{n+1}(M_n+1) > p_{n+1}(M_n+2)$, then
$(p_{n+1}(0), p_{n+1}(1), \ldots, p_{n+1}(2{n+2}))$ is unimodal with
mode $M_n+1$.
If $p_{n+1}(M_n) > p_{n+1}(M_n+1) > p_{n+1}(M_n+2)$, then
$(p_{n+1}(0), p_{n+1}(1), \ldots, p_{n+1}(2{n+2}))$ is unimodal with
mode $M_n$.
All that remains to prove unimodality of $(p_{n+1}(0), p_{n+1}(1), \ldots, p_{n+1}(2{n+2}))$ is to show that the case
$$p_{n+1}(M_n) < p_{n+1}(M_n+1) > p_{n+1}(M_n+2)$$
cannot occur. Compare to the OP's $p > q < r$ is not allowed.