The null hypothesis of ANOVA is that the within-group variances are equal and all group means are equal. Under this hypothesis, your data provide enough information for testing, provided you can justify assuming the sampling distributions of the group means are close to Normal.
Here's a brief analysis. To establish notation, let there be $d\ge 2$ groups of sizes $n_1, n_2, \ldots, n_d$ comprising $N=n_1+\cdots + n_d$ independent observations. Let the common variance be $\sigma^2$ and the common mean be $\mu.$ If $\mathcal{L}$ is the likelihood, write $\Lambda = -2\log(\mathcal L),$ which is minimized when the likelihood is maximized.
These assumptions imply the group means $x_i$ have Normal$(\mu, \sigma^2/n_i)$ distributions. Therefore
$$\Lambda = \sum_{i=1}^d \log\left(\frac{\sigma^2}{n_i}\right) + \frac{(x_i-\mu)^2}{\sigma^2/n_i}.$$
Critical values of the gradient of $\Lambda$ produce the familiar estimates
$$\hat \mu = \frac{1}{N}\sum_{i=1}^d n_i x_i$$
and
$$\hat\sigma^2 = \frac{1}{d}\sum_{i=1}^d n_i(x_i-\hat\mu)^2.$$
At this point it is natural to re-express the data in terms of Z scores as
$$z_i = \frac{x_i - \hat\mu}{\sqrt{\hat\sigma^2 / n_i}}$$
because each of these is approximately standard Normal and they are approximately independent. Consequently, you could inspect this set of Z scores for deviations from a standard Normal distribution. You might use a Normal-theory outlier test, for instance; or you could construct a statistic such as the Kolmogorov-Smirnov statistic or Anderson-Darling statistic. For formal testing, bootstrapping the distribution of this statistic from the estimates will work (and helps us avoid extensive further analysis!).
As an example, the estimates for the data $X=(6.58, 7.4, 3.2)$ with group sizes $n=(20, 2, 15)$ are
$$(\hat \mu, \hat\sigma^2) = (5.25, 35.89)$$
with corresponding Z scores
$$z = (0.99, 0.51, -1.33).$$
These aren't unusual, so we haven't found significant evidence of a difference in group means. Indeed, it's scarcely possible to do so with just three groups--one of the means would have to be far from the other two.
Suppose, though, that you have more data from more studies and therefore you have means of more than three groups. Then this approach could yield significant results. As an example, consider ten groups with means $x=(1,2,\ldots, 9, 30).$ That last value of $30$ is extreme. The estimated parameters are now $\hat\mu=7.5$ and $\hat\sigma^2=124.5$ with Z scores
$$z = (-0.82, -0.70, \ldots, 0.06, 0.19, 2.85).$$
That last one ($2.85$) is a little unusual and indeed, the parametric boostrap of the KS statistic gives a p-value of $0.011,$ indicating a significant difference.
The following R code gives an implementation. The p-value for the data in the question is $0.605.$
#
# Data.
#
x <- c(6.58, 7.4, 3.2)
n <- c(20, 2, 15)
# x <- c(1:9, 30)
# n <- rep(2, length(x))
#
# Parameter estimates.
#
theta.hat <- function(x, n) {
N <- sum(n)
m <- sum(n*x) / N
s2 <- mean(n*(x-m)^2)
c(x.hat=m, s2.hat=s2)
}
(theta <- theta.hat(x, n))
#
# Bootstrap statistic.
#
KS.stat <- function(x, theta, n) {
max(abs((1:length(x))/(length(x)+1) - sort(pnorm(x, theta["x.hat"], sqrt(theta["s2.hat"]/n)))))
}
stat <- KS.stat(x, theta, n)
#
# Display Z scores.
#
print(signif(x - theta["x.hat"]) / sqrt(theta["s2.hat"] / n), 2)
#
# Bootstrap the statistic.
#
sim <- replicate(1e4, {
y <- rnorm(length(n), theta["x.hat"], sqrt(theta["s2.hat"]/n))
p <- theta.hat(y, n)
KS.stat(y, p, n)
})
print(c(`p-value`=mean(c(stat, sim) >= stat)))
