The two tests (logistic regression and chi-square) are equivalent and a power analysis should give the same answer.
You are assuming that a value of 0.15 for f2 and w are the same effect size, they're not. A small value of w is 0.1, a small value of f2 is 0.02.
Edit: Elaborated on the similarity of the two approaches.
IF you give the same data to logistic regression and a chi-square test (strictly: without Yates' correction), you get the same result. Here's an example
> x <- rbinom(100, 1, 0.2)
> y <- rbinom(100, 1, 0.2)
> chisq.test(table(x, y), correct=FALSE)
Pearson's Chi-squared test #'
data: table(x, y)
X-squared = 0.155, df = 1, p-value = **0.694**
In chisq.test(table(x, y), correct = FALSE) :
Chi-squared approximation may be incorrect
> summary(glm(y ~ x, family="binomial"))
glm(formula = y ~ x, family = "binomial")
Min 1Q Median 3Q Max
-0.753 -0.753 -0.753 -0.668 1.794
Estimate Std. Error z value Pr(>|z|)
(Intercept) -1.114 0.251 -4.43 9.4e-06 ***
x -0.272 0.693 -0.39 **0.69**
Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
(Dispersion parameter for binomial family taken to be 1)
Null deviance: 110.22 on 99 degrees of freedom
Residual deviance: 110.06 on 98 degrees of freedom
Number of Fisher Scoring iterations: 4
The p-values are the same, so the power should be the same. I can't remember the formulas for the two different versions of the effect size. Effect size measures are a little weird because in the old days you wanted to minimize the number of tables that you put into books (so we have, for example, $f^2$ instead of $R^2$, when there's a direct relationship between them, and $R^2$ is what everyone understands).