Assume first the radius of the ball is $r=1$.
Write your point $X$ as a scaled point of the surface of the unit sphere: $X=R\sigma$, where $R=\|X\|$ and $\sigma=X/R$. It should be easy to convince yourself (using spherical symmetry) that $R$ and $\sigma$ are independent, and that what you want is $E[R^2]E[\sigma_1^2]$. The density function for $R$ is $n\rho^{n-1}$, and $E[\sigma_1^2] = E[Z_1^2/(Z_1^2+\cdots+Z_n^2)]= 1/n$, where the $Z_i$ are iid $N(0,1)$ variates. Thus, $$\text{Var}(X_1) = E[X_1^2] -(E[X_1])^2 = E[X_1^2]-0^2 = \int_0^1 n\rho^{n-1} \rho^2\,d\rho \times \frac 1 n = \frac {n}{n+2} \times\frac 1 n = \frac 1 {n+2}.$$
Relaxing the assumption that $r=1$, a simple rescaling give the general expression that $\text{Var}(X_1)=\frac{r^2}{n+2}$. (Because if
$X$ is uniformly distributed on the $1$-ball then $rX$ is uniformly distributed on the $r$-ball and
for any random variable $Y$, we always know $\text{Var}(rY) = r^2\text{Var}(Y)$.)
Note we distinguish here between the non-random $r$, the radius of the problem-statement ball; $R$ the length of the random point in the ball; and $\rho$, the
variable of integration when taking the expectation of $R^2$.
Going beyond the question asked here, this formulation can help in evaluating higher moments $E\|X_1\|^k$ for $k>2$, at the price of added complexity. One has $E\|X_1\|^k=r^k E[R^k]E[\sigma_1^k]$. The first expectation is just $\int_0^1n\rho^{n-1}\rho^k d\rho$ but the second involves the ``Beta'' distribution, and is given by a ratio of gamma functions.