Test statistics and p-values are random; they don't really estimate anything.
Under a point-null hypothesis the p-value (at least for a continuously distributed test statistic) should be uniformly distributed between 0 and 1.
[For discrete test statistics their cdf is a step function which touches the $y=x$ line at each jump, but the jumps might not be all the same height - they usually aren't.]
You seem to be expecting that the p-value and the chi-squared value should converge toward some particular value at large sample sizes (whence calculating a standard error might make sense). This is not the case.
To answer the direct question, the variance of the distribution of a p-value under the null (for a continuous test statistic) should be $1/12$. The standard deviation of the distribution of the p-value should then be the square root of that. Similarly, the variance of a chi-squared($k$) is $2k$, so the standard deviation is $\sqrt{2k}$.
[Under the alternative, p-values may tend to be typically small but are generally quite right skew. Under the alternative, the test statistic simply tends to be stochastically larger than under the null -- it is also not converging to some particular value. In the case of chi-square tests, typically under the alternative the test statistic has a noncentral chi-squared distribution, or approximately so. This may be the case for the tests you're looking at but I have not checked that it's so.]
I don't think there's much value in pursuing this line of thought, the distribution of the p-value isn't converging toward nor gathering about anything but 0, and then only under the alternative. The standard deviation of the distribution of the p-value under the null doesn't tell you anything; it's the same for every continuous test statistic (with a point null).
If you want to talk about accuracy in some useful sense then you should probably be looking at some kind of measure of effect size, or some confidence interval for whatever quantity your test is trying to pick up.
