Is it possible to create a completely random integer between 1 and 13 using standard dice in a D&D dice kit? I am asking if it is possible to choose a random integer using a certain set of dice. Assume for these calculations that the dice are perfectly randomly distributed. There are 7 dice in a set, with the following possible outcomes for each dice:


  2, 4, 6, 8, 10, 12, and 20


You are free to interpret the outcomes as any value you like.

The dice can be rolled in any order, combination, or quantity desired.

The objective is to have 13 possible outcomes, perfectly random and uniformly distributed.
This is impossible if you require that there is a fixed upper bound on the number of possible rolls: in this case, the only probabilities you can get are of the form $\frac{n}{m}$ where $m$ is a product of the numbers $2, 4, 6, 8, 10, 12, 20$, and in particular cannot be divisible by $13$. 

This is straightforward if you don't require such a fixed upper bound: as mentioned in the comments, you can discard and reroll. This strategy, like any successful strategy, necessarily has the property that it is possible that you'll have to reroll arbitrarily many times. But the probability is exponentially decaying so it's not really an issue. 

Edit: An interesting follow-up question might be to find a strategy that minimizes the expected number of rolls. Suppose you have a strategy involving $n$ rolls per iteration, with probability $p$ of succeeding each iteration (and probability $1 - p$ of needing to reroll). Then the expected number of rolls turns out to be $E = \frac{n}{p}$. So, comparing the strategies that have been suggested so far:


Travis's $4$ and $10$ strategy has $n = 2, p = \frac{39}{40}$, so $E = \frac{80}{39} \approx 2.05$.
Amitai's $20$ strategy has $n = 1, p = \frac{13}{20}$, so $E = \frac{20}{13} \approx 1.54$.
M. Wind's "all the dice" strategy has $n = 7, p = \frac{921596}{921600}$, so $E = \frac{6451200}{921596} \approx 7.00$. 


Amitai's strategy is in fact the only possible strategy with an expected number of rolls of less than $2$ (so I take back what I said about it requiring more rolls than Travis's!), although one might go on to calculate and compare the variance of the number of rolls as well...
Try rolling a D10 and a D6 with the D6 values being:

1-2 = 1
3-4 = 2
5-6 = 3