Consider a sequence of $2k$ dice each with the possible values $1, 2, \ldots, 2n.$ (The question concerns $k=n=3.$) The possible pairs are $\{1,n\},$ $\{2,2n-1\},$ and so on, through $\{n,n+1\}.$ Denote such a pair by its smallest value $i$ and for each $i$ let $k_i\gt 0$ be the number of such pairs that can be located in the sequence. We need to count the number of equally probable sequences for which $k_1+k_2+\cdots + k_n = k.$
Such a sequence is determined by (a) which pairs occur in it and (b) where in the sequence of $2k$ values each such pair occurs. The permutation group on $2k$ elements acts on the set of such sequences. For all $i,$ the stabilizer of such a sequence permutes the $k_i$ values of $i$ among themselves and the $k_i$ values of $2n+1-i$ among themselves. Thus, applying the method described at https://stats.stackexchange.com/a/415878/919, the number of ways of producing a sequence designated by $\mathrm{k}=(k_1,k_2,\ldots,k_n)$ is
$$p(\mathrm{k}) = \frac{(2k)!}{(k_1!)^2(k_2!)^2\cdots(k_n!)^2}.$$
Thus, the chance is obtained by summing $p(\mathrm{k})$ over all possible $\mathrm{k}$ whose components sum to $k$ and dividing by the total number of sequences, $(2n)^{2k}.$
These possibilities correspond to the weak compositions of $k$ into $n$ parts, which number $\binom{k+n-1}{n-1}.$ However, the amount of calculation is smaller than this, because $p(\mathrm{k})$ does not depend on the order of the $k_i.$ We may therefore do the calculation for all $k_1\ge k_2\ge \cdots \ge k_n \ge 0$ (giving a partition of $k$), multiplying each by the number of distinct re-orderings of $\mathrm k.$ Such sequences correspond to the Ferrers diagrams for $n$ having at most $k$ rows. They are relatively easy to enumerate.
With $k=n=3,$ we have $\binom{3+3-1}{3-1}=10$ possibilities for $\mathrm k,$ but they fall into just three groups corresponding to the partitions $3 = 2+1 = 1+1+1:$
$$p(3,0,0)=p(0,3,0)=p(0,0,3) = \frac{6!}{3!^2} = 20;$$
$$p(2,1,0)=p(2,0,1)=p(1,2,0)=p(1,0,2)=p(0,2,1)=p(0,1,2) = \frac{6!}{2!^21!^2}= 180;$$
$$p(1,1,1) = \frac{6!}{1!^21!^21!^2} = 720.$$
The answer therefore is
$$\Pr(\text{three pairs}) = \frac{3\times 20 + 6\times 180 + 1\times 720}{6^6} = \frac{(5)(31)}{(2^4)(3^5)} \approx 3.9866\%.$$
You have likely figured out by now why I limited this analysis at the outset to dice with even numbers of sides: it eliminates the possibility that some values would be paired with themselves. (For instance, on a five-sided die with values 1 through 5, 3+3=6 forms a pair.) A similar analysis can be carried out for the odd-sided dice. It results in a slightly more complicated formula due to that self-pairing possibility.
