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.

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, whatever the real distribution of $x_{permuted}$ is. 

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.

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.

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.