Generally, I agree with Jeromy's arguments that the mean is a reasonable statistic for Likert scales.
What could speak for the median, is that the median is a much more robust measure of location as it protects against outliers (it has the highest possible breakdown point of 50%). However, as Likert scales are bounded scales, the possibility of extreme outliers is very low (only if your data is extremely skewed). Furthermore, the median usually trims too much from the data, so you could consider using trimmed means instead. An amount of 20% trimming usually is recommended .
If you want to calculate a paired test of the difference of medians, I would recommend to compare the means using a percentile bootstrap method (this is the only method for comparing medians that works well in the case of tied values, see Wilcox, 2005 ).
In the WRS package for R, there is a function called
trimpb2 which does this calculation for two independent samples (you can also calculate a p value for trimmend means with that function). In your case, however, you need to compare dependent groups. In this case, you can also do a bias-adjusted percentile bootstrap method .
Note, however, that the difference of the medians of the marginal distributions is not the same as looking at the median of the difference scores. The first answers the question 'How does the typical response from the first group differ from the second' and is performed by the WRS function
rmmcppb. The second answers the question 'What is the typical difference score' and is performed by the WRS function
 Wilcox, R. R. (2005). Introduction to robust estimation and hypothesis testing. San Diego: Academic Press.
 Wilcox, R. R. (2006). Pairwise comparisons of dependent groups based on medians. Computational Statistics & Data Analysis, 50, 2933-2941. doi:10.1016/j.csda.2005.04.017