$\textbf{Answer}$

$\frac{13}{36}$.

$\textbf{Analysis}$

Assuming two dice have $m$ and $n$ different numbers, respectively. Then they can roll up to $(m\times n)$ different numbers. Given the pair rolls every number from $2$ to $12$, $(m\times n)$ must be no less than $11$. It follows that at least one of them must have no less than $4$ distinct numbers. (Otherwise, if both dices have up to $3$ distinct numbers, then they can only roll up to $3\times 3=9$ different numbers.) Let this die be the first die. Then the number of distinct numbers it has can only be $6$, $5$, or $4$.

If the first die has six different numbers, for every number on the second die, there is at most one number on the first die which can make their sum equal $7$. This means that the overall probability is at most $\frac{1}{6}$.

If the first die has five different numbers with one duplicate, then the second one must have at least three different numbers because $m\times n \ge 11$. With two dice, there are totally $6\times 6=36$ possible pairs which is a fixed number. Therefore, finding the maximum probability will be the same as finding the maximum number of pairs whose sums equal $7$. Given the occurrence of these five different numbers on the first dice are $2$, $1$, $1$, $1$, $1$ and each of these number has just one matching number to sum to $7$, it can be shown that the maximum will be achieved when the second dice has three different numbers with occurrences as $4$, $1$, $1$ and the matching numbers are paired in these orders. In this case, the probability will be $$\frac{2\times 4 + 1\times 1 + 1\times 1}{6\times 6}=\frac{10}{36}$$

If the first dice has four different numbers, the occurrences of these numbers can be either $(3, 1, 1, 1)$ or $(2, 2, 1, 1)$. Accordingly, the second dice should have at least three different numbers because $m\times n \ge 11$. Applying the same logic mentioned in the previous case will find the maximum probability can be archived as $$\frac{3\times 4 + 1\times 1 + 1\times 1}{6\times 6}=\frac{14}{36}$$

Combining these three cases gives the largest theoretical probability as $\frac{14}{36}$. However, this value is un-achievable because it will conflict with the given condition that these two dice can roll every number from $2$ to $12$ (see the note below). The next largest possible value is $\frac{13}{36}$. This can be achieved by the following two dice: $\{1, 1, 1, 1, 3, 5\}$ and $\{1, 2, 6, 6, 6, 7\}$. Hence, the final answer is $\frac{13}{36}$.

$\textbf{Note}$

When a problem asks for the maximum or minimum values, a usual approach is to first seek a theoretical limit and then find a solution satisfying the theoretical limit. This example shows the importance of the second step. In this example, it can be proved that $14$ combination is impossible, though the proof is a bit lengthy. We will skip it here.