Connection between max and min of symmetric random variables

The claim isn't even true for a single variable. Let the sample space be ${0, 1}$ with $P(0) = 1/3$, $P(1) = 2/3$ and let $X_1(0) = -2$, $X_1(1) = 1$. Then $mathbb{E}[X_1] = 0$ but $P(X_1 > 0) neq P(X_1 < 0)$.

Edit: because I slightly misread the question, I originally gave a counter example which didn't actually work. I have removed it and replaced it.

Let's consider a stronger version of "symmetric"; that for each $i$, the random variable $X_i$ hasn't the same probability distribution as $-X_i$ does. This actually still isn't enough. For consider random variables $X_1, X_2, X_3$ defined as follows:

There is a 1/3 chance that $X_1 = X_2 = X_3 = 1$. For the other $2/3$ of the time, independently assign each $X_i$ to $0$ or $-1$ with equal probability.

Formally, the sample space is ${(1,1,1)} cup {0, -1}^3$ with probability distribution $P(x) = 1/3$ if $x = (1,1,1)$ and $1/12$ otherwise. The variables $X_1, X_2, X_3$ are the projection functions.

Clearly, each $X_i$ has a $1/3$ chance of being each of $1$, $0$, and $-1$ and is therefore symmetric under this stronger definition. The probability of the maximum value being negative is only $frac{2}{3} (frac{1}{2})^3 = frac{1}{12}$ since this only occurs when $X_1 = X_2 = X_3 = -1$, while the probability of the minimum being positive is $frac{1}{3}$.