I have data of Likert Scale ratings (1 to 7), and I had originally planned to conduct a paired t-test on until I realized it wasn't normally distributed. I went down the path of using Wilcoxon Signed-Rank Test except that everywhere I read about it they say that it should only be conducted on medians (not the means). The problem here is that when I compute Wilcoxon Signed-Rank Test on the medians, the majority of participants are excluded since the ranks cannot be differentiated since all participants fall into a 0, 1, and 2 Likert point difference across the conditions. I don't encounter this problem if I conducted it on means of the data.

So I am wondering is what is the problem with conducting Wilcoxon Signed-Rank Test on the means? More generally, if I have non-normal ordinal data, is there a different statistical test I ought to be using?
First get the terminology right: you are considering Likert-type items (not Likert scales which are sum scales of Likert-type items).

My first test would be chi-square of conditions crossed with Likert-type items; if nonsignificant, you're done. But if you find a significant result, so you reject independence, you need to address what is different between conditions: measures of central tendency (mode, median, mean) and/or measures of dispersion (range, variance). Typically, researchers are interested in means, and the associated tests such t or F. The problem of these test is not normality, as these statistics been shown to be fairly robust against deviations of normality in numerous studies. The issue is that if you find a significant difference of say, 0.3141527 between groups, what that means in real terms. The median test will give you the medians of the groups and the assessment whether observed differences are significant - and since the medians are expressed in term of the used scale they have an unambiguous interpretation.
You don't need the data to be normally distributed; you only need the sampling distribution of the means to be roughly approximated by a normal distribution. With data like yours, it is, assuming your sample size is more than about 3. You can use a paired t test without apology.