This is quite straightforward, there is no need to infer a distribution under the Null Hypothesis. Your p-value is just the number of times $x_{permuted}$ get superior or equal to $0.5$, divided by the number of permutations made.    
This fits the definition of the p-value: *"If H0 is true and a new sample is drawn, what is the probability to get at least such extreme results ?"*

Maybe you mixed things up with the Bootstrap procedure which is generally done to estimate the "*real*" distribution of your statistic of interest. 

I don't say your approach is completely wrong, if your distribution looks like normal, you could eventually do a `z-test`, and it should give a quite reliable p-value. But I think the spirit of the permutation test is just to count how often you indeed get equal or more extreme results because you have a direct access to it. For the sake of the comparison it would be interesting that you give how often $x_{permuted}$ get superior or equal to $0.5$ in your data set, we could compare it with what would give a 1-tailed z-test.