Edit: The "second moment method" you've stated is false, as shown for example by $P(X=1)=p$ and $P(X=0)=1-p$ with $p>1/2$. See this Wikipedia article for a discussion of the more complicated inequalities sometimes called the first or second moment methods.
But assuming I'm right that you're interested in the probabilistic method for existence proofs (which is hard to tell since you didn't give any context), it's not really the right question to ask. The first and second moment methods have names at all because they're frequently useful and easy to carry out. I don't know offhand of applications where you can't use those but can use a higher moment method. Instead one typically has to turn to more sophisticated tools. Check out "The Probabilistic Method" by Alon and Spencer for a taste of lots of those.
Added: Okay, here's the answer to your revised question: $$P(\vert X - \mathbb{E}X \vert \ge (\mathbb{E}\vert X- \mathbb{E} X \vert^n)^{1/n} ) > 0,$$ and when $n$ is even you cannot replace $(\mathbb{E}\vert X- \mathbb{E} X \vert^n)^{1/n}$ with any larger quantity depending only on the first $n$ moments of $X$. To see the latter claim, let $Y=X-\mathbb{E}X$ and observe that knowledge of the first $n$ moments of $X$ is equivalent to knowledge of $\mathbb{E}X$ and the first $n$ moments of $Y$. My claim is that there is a random variable $Y$ with $\mathbb{E}Y=0$ such that $$P(Y^n > \mathbb{E} Y^n) = 0,$$ and indeed this is true if $P(Y=-1)=P(Y=1)=1/2$.