Cohen's d is a measure of the standardized difference in means.
(mean($X_1$)-mean ($X_2$))/sigma
So the null hypothesis tests whether this standardized difference is equal to zero. This is different from the original null hypothesis which tests whether the non-standardized difference in means is equal to zero.
It might help to see how Cohen's d can be calculated using t. You can find the details in the compute.es package documentation (http://cran.r-project.org/web/packages/compute.es/compute.es.pdf):
t = d$*$sqrt($n_1$*$n_2$/($n_1$ + $n_2$))
rearranging will give you d.
[Addition to the above:]
The d = 1.67 tells you that the difference between the two groups is about one and two-thirds of a standard deviation. The original p=0.05657 was calculated for the non-standardized difference in means. The t-statistics for this difference follows a 'central t-distribution' - that is, it is symmetric around 0. 'Central t-distributions' have one parameter: the degrees of freedom.
The difference with the effect size is that the t-statistic for it is non-centrally distributed (it is not symmetric around 0). 'Non-central t-distributions' have two parameters: the degrees of freedom and a 'non-centrality' parameter. Rather than getting into further details, you can find a straightforward introduction here:
Cumming, G. & Finch, S. (2001) A primer on the understanding, use, and calculation of confidence intervals that are based on central and noncentral distributions. Educational and Psychological Measurement, 61, 633-649.
They give a readable account of why the effect size has a non-central t-distribution on pages 549-551.
Also: you asked for an explanation of the p-value. P-values depend on the distribution of the test statistic which I've addressed in the above (the details of which are in the reference). Hopefully that helps! If you want an explanation of what a p-value is generally, then likely that would require a different post and has probably been asked elsewhere.