Question: Assume that $X\_n$ is a sequence of a zero-mean and unit variance random variables (and maybe having density w.r.t. to Lebesgue). Can we conclude that $ P(X\_n \in [0,R\_n]) $ is bounded away from zero eventually, say $$\liminf\_{n \to \infty} P(X\_n \in [0,R\_n]) > 0$$ assuming that $R\_n \to \infty$. Intuitively, $X\_n$ should put some nonvanishing amount of mass on the positive real line because of the zero-mean assumption and the condition of unit variance should prevent that mass from escaping to infinity (?)

Here is a partial argument: We have
\begin{align\*}
P(X\_n \in [0,R\_n]) &= 1 - P(X\_n > R\_n) - P(X\_n < 0) \\
&\ge 1 - P(|X\_n| > R\_n) - P(X\_n < 0) \\
&\ge 1 - \frac{E X\_n^2}{R\_n^2} - P(X\_n < 0) \\
&= 1 - o(1) - P(X\_n < 0)
\end{align\*}
Thus, the problem reduces to arguing that $P(X\_n < 0)$ stays away from 1.

Answer: 
Here's a proof if a third moment condition is satisfied. 

Suppose that $\mathbb EX=0$, $\mathbb EX^2=1$ and $\mathbb E|X|^3\le K$. 
Then let $Z=|X|$. We use Cauchy-Schwarz: $Z^2=Z^{1/2}Z^{3/2}$, so that $(\mathbb EZ^2)^2\le \mathbb EZ\cdot \mathbb EZ^3$. This gives $\mathbb EZ\ge \frac 1K$. Hence $\mathbb EX\mathbf 1\_{X>0}\ge \frac{1}{2K}$. 

Also $\mathbb EX^2\mathbf 1\_{X>0}\le 1$. Now $\mathbb EX\mathbf 1\_{X>4K}\le (1/4K)\mathbb EX^2\mathbf 1\_{X>4K}\le 1/(4K)$, so that $\mathbb EX\mathbf 1\_{X\in [0,4K]}\ge 1/(4K)$ and $\mathbb P(X\in [0,4K])\ge 1/(16K^2)$.
A standard technique to lower-bound $\mathbb P(0 < X < r)$ as a function of $\mathbb E[|X|^3]$ subject to $\mathbb E[X] = 0$ and $\mathbb E[X^2] = 1$
is to consider a suitable linear combination of $|X|^3$, $X$, $X^2$ and $1$ that is negative on $(0,r)$ and nonnegative outside that interval and has a negative expected value. In this case try
$$ f(x) = |x|^3 - b r x^2 - (1-b) r^2 x$$ 
For $x \ge 0$ we have $f(x) = x (x-r) (x-(b-1)r)$, so this satisfies the sign requirements there if $b < 1$.
$f(x) > 0$ for $x < 0$ if $b^2 + 4 b - 4 < 0$, which is true if $0 \le b \le 2\sqrt{2}-2$.
We have $\mathbb E[f(X)] = \mathbb E[|X|^3] - b r$, which we want to be negative. The minimum value of $f(x)$ on $[0,r]$ turns out to be
$$ v = - \dfrac{r^3}{27} \left( 2 (b^2 - 3 b + 3)^{3/2} + 2 b^3 - 9 b^2 + 9 b\right)$$
The conclusion is then that if $\mathbb E[|X|^3] < b r$ where $0 < b \le 2\sqrt{2}-2$, 
$$ \mathbb P(0 < X < r) \ge \frac{ br - \mathbb E[|X|^3]}{-v} $$ 
Taking $b = 2\sqrt{2}-2$, 
we get $$\mathbb P(0 < X < t E[|X|^3]) \ge c \dfrac{t-(1+\sqrt{2})/2}{t^3 \mathbb E[|X|^3]^2}$$
for $t > (1+\sqrt{2})/2$, where $c \approx 4.43035992$.