An attempt to provide the whole distribution using analytic combinatorics (though one answer is already accepted). In particular, we can obtain the first and the second moment using this technique (and maybe some more by induction if you want a limiting distribution and not only first two moments).

The generating function for Catalan paths
$$
Cat(z) = \sum_{n \geq 0} c_n z^n
$$
satisfies an equation
$$
Cat(z) = z^2 Cat(z) + 1 \enspace , \quad
Cat(z) = \dfrac{1 - \sqrt{1 - 4z^2}}{2z^2}
$$
where 1 corresponds to an empty path, then $z Cat(z) \cdot z$ corresponds to $ \nearrow$ times Catalan path times $\searrow$, then another Catalan path $Cat(z)$. The idea is to mark the area inside Catalan path, i.e. consider a bivariate generating function
$$
C(z,u) = \sum_{n, k \geq 0} c_{n,k}z^n u^k \enspace ,
$$
where $ c_{n,k}$ is equal to the number of Catalan paths of semilength $n$ and area $k$. An equation here is the following one:
$$
C(z,u) = z^2 u C(zu^2, u)C(z,u) + 1
$$
Explanation: the first part which starts from $\nearrow \ldots \searrow$ is a Catalan path of area $k + 2n + 1$, where $k$ is the area of upper part, $n$ is the semilength and $1$ is because you glue two parts of a triangle. For that reason,
$$
uC(zu^2,u) = u\sum_{n, k \geq 0} c_{n,k}(zu^2)^n u^k
= \sum_{n,k \geq 0} c_{n,k} z^n u^{k + 2n + 1} \enspace .
$$
The second Catalan part is glued without chantings, i.e. we count already existing area as an additional summand.

Next step is to obtain the moments from the functional equation. Though the equation cannot be solved explicitly, we can still obtain some information.
Recall that k-th factorial moment is given by
$$
\dfrac{[z^n] \partial_u^k C(z,u) |_{u=1}}{[z^n] Cat(z)}
$$
I denote $ C^\square(z) = \partial_u C(z,u)|_{u=1} $, $ C^{\square\square}(z) = \partial^2_u C(z,u)|_{u=1} $ for brevity. Let's take a derivative then:
$$
C^\square(z) = z^2 C^2(z) + z^2 C^\square(z) C(z) + z^2(2C_z + C^\square)C(z)
\enspace ,
$$
where $ C_z(z) := \partial_z C(z) $. Thus, $C^\square(z)$ can be explicitly expressed through $C(z) $ and $\partial_z C(z)$, and that's how you can obtain (already present) result on expectation of the area.

Then you can repeat and obtain explicit expression for the second derivative, thus having a bound for the variance. In principle, you can try to derive some general pattern on the moments of this distribution (or more concretely, on the singularity $\rho(u)$ and nature of singularity of complex function $C^{k\square}(z)$), and apply Theorem IX.8 from Analytic Combinatorics of Flajolet and Sedewick. I believe that some of the summands in the functional equation will be negligible which will simplify the analysis.

UPD: Actually, it happens that corresponding distribution is of area Airy type. See the article "Analytic Variations on the Airy Distribution" by Flajolet and Louchard:
http://algo.inria.fr/flajolet/Publications/FlLo01.pdf

Bernoulli excursionis another name (more popular among probabilists) for Catalan path. Look at A Bernoulli Excursion and Its Various Applications by Takács for quantitative results on the area statistic of the Bernoulli excursion as well as its scaling limit, the (normalized) Brownian excursion. $\endgroup$ – HMPanzo May 13 '17 at 3:53