Eleven.
For $s$ odd let $S_s$ be the set of tuples $(a_1,\ldots, a_k)$ with $k\ge 1$, $a_k\in\{1,3,5,7,9,11\}$ and all other $a_i\in\{2,4,6,8,10\}$ and $a_1+\ldots +a_k=s$. Define $|(a_1,\ldots,a_k)|:=k$.
The probability of ending at sum $s$ is
$$P(s) =\sum_{\alpha\in S_s}12^{-|\alpha|}.$$
If $s<11$ we define an injection $f\colon S_s\to S_{11}$ by
$$(a_1,\ldots,a_k)\mapsto (a_1,\ldots,a_{k-1},a_k+2)$$
(noting that $s<11$ implies $a_k<11$).
Since $(2,2,2,2,2,1)$ is not in the image of this map and $|f(\alpha)|=|\alpha|$ we conclude $P(11)>P(s)$.
If $s>11$ then observe that
$$g\colon S_s\to\bigcup_{k=1}^6S_{s-2k},\\
(a_1,\ldots,a_m)\mapsto(a_1,\ldots,a_{m-1}-1)$$
is a bijection. As $|g(\alpha)|=|\alpha|-1$, we have
$$ P(s)=\frac1{12}\sum_{k=1}^6 P(s-2k)$$
and by induction $P(s)\le \frac12P(11)$ for $s>11$.
By the binomial distribution, (easy to find online) the probability that $k$ seeds generate is ${6 \choose k}(0.4)^k(1-0.4)^{6-k}$. The maximum will always occur either left or right of the expected value of the distribution, which is $6 \cdot 0.4 = 2.4$. So you just need to compute the binomial distribution probability values for $k=2$ and $k=3$ to find the maximum.
Best Answer
Note that there are a total of $6 \times 6 \times 6 = 216$ options. The possible sums are from $3$ to $18$.
Also note that the distribution has to be symmetric since if a roll gives $x,y,z$ adding to $n$, then $7-x,7-y,7-z$ adds to $21-n$.
Hence, the number of ways of getting $n$ is same as the number of ways of getting $21-n$. We would hence expect the maximum to occur for $10$ and $11$.
Number of ways to get $3$ is $1$ i.e. $\dbinom{2}{2}$, which is the same as the number of ways to get $21-3=18$.
Number of ways to get $4$ is $3$ i.e. $\dbinom{3}{2}$, which is the same as the number of ways to get $21-4=17$.
Number of ways to get $5$ is $6$ i.e. $\dbinom{4}{2}$, which is the same as the number of ways to get $21-5=16$.
Number of ways to get $6$ is $10$ i.e. $\dbinom{5}{2}$, which is the same as the number of ways to get $21-6=15$.
Number of ways to get $7$ is $15$ i.e. $\dbinom{6}{2}$, which is the same as the number of ways to get $21-7=14$.
Number of ways to get $8$ is $21$ i.e. $\dbinom{7}{2}$, which is the same as the number of ways to get $21-8=13$.
Number of ways to get $9$ is $25$ i.e. $\dbinom{8}{2}-3$, which is the same as the number of ways to get $21-9=12$.
Number of ways to get $10$ is $27$ i.e. $\dbinom{9}{2} - 3 - 6$, which is the same as the number of ways to get $21-10=11$.
Now by symmetry you can get the number of ways to get the sum between $11$ and $18$.
Hence, the distribution peaks at $10$ and $11$ as expected.
As a sanity check, we have $2 \left( 1 + 3 + 6 +10 +15 + 21 + 25 + 27\right) = 216$.
Below is the distribution of the number of times a number occurs as a sum of three dices, where the $X$-axis the sum of the three dice and $Y$-axis is the number of times the sum occurs.