To keep things simple we will do some induction:
Set $A_r = \{(x_1, \dots, x_r) \in \mathbb R^r : \forall i \in \{1,\dots, n\}: \vert x_i \vert \leq 1\}$. Consider the integral
$\int_{A_r} 1 \ d\lambda(x_1, \dots, x_r)$, we will prove by induction that
$$\int_{A_r} 1 \ d\lambda(x_1, \dots, x_r) = 2^r.$$
For $n = 1$ we get
$$\int_{A_1} 1 \ d x_1 = \int_{-1}^1 1\ dx_1 = 2.$$
Now let the statement hold for $r \in \mathbb N$. Then we get for $r + 1$ with Fubini that
$$\int_{A_{r+1}} 1 \ d\lambda(x_1, \dots, x_{r+1}) = \int_{-1}^1 \bigg(\int_{A_r} 1 \ d\lambda(x_1, \dots, x_r) \bigg)dx_{r+1} = \int_{-1}^1 2^r\ dx_{r+1} = 2^{r +1}.$$
Now to the second part. Let $k := n - r$ and $B_{k} = \{(x_{r+1}, \dots, x_n) \in \mathbb R^{n-r} : x_{r+1}^2 + x_{r+2}^2 \leq 1, \ldots, x_{n-1}^2 + x_{n}^2 \leq 1 \}$ for even $k$. Now we prove by induction that
$$\int_{B_k} 1 \ d\lambda(x_{r+1}, \dots, x_n) = \pi^{k/2}.$$
For $k = 2$ we get by using polar coordinates
$$ \int_{B_2} 1\ d\lambda(x_{r+1}, x_{r+2}) = \int_0^1 \int_0^{2 \pi} r\ d\varphi dr = \int_0^1 2\pi r\ dr = \pi.$$
Now let the statement hold for $k \in \mathbb N$ even. Then we get for $k + 2$ with Fubini that
\begin{align*}\int_{B_{k+2}} 1\ d\lambda(x_{r+1}, \dots, x_{n + 2}) &= \int_{B_2} \bigg( \int_{B_{k}} 1\ d\lambda(x_{r+1}, \dots, x_n)\bigg) d\lambda(x_{n+1}, x_{n + 2}) = \int_{B_2} \pi^{k/2}\ d\lambda(x_{n+1}, x_{n + 2}) \\ &= \pi^{k/2} \int_0^1 \int_0^{2 \pi} r\ d\varphi dr = \pi^{k/2} \pi = \pi^{(k + 1)/2}\end{align*}
All together we can get know with Fubini
\begin{align*}\operatorname{vol}(A) &= \int_A 1\ d\lambda(x_1, \dots, x_n) = \int_{A_r} \bigg(\int_{B_k} 1\ d\lambda(x_{r+1}, \dots, x_n) \bigg) d\lambda(x_1, \dots, x_r) \\ &= \int_{A_r} \pi^{k/2}\ d\lambda(x_1, \dots, x_r) = \pi^{k/2} \int_{A_r}1 \ d\lambda(x_1, \dots, x_r) = \pi^{k/2} 2^r.\end{align*}
And that was the result we were aiming for.
The "physics" version:
Denote the unit disk by $D = \{x \in R^2 : x_1^2 + x_2^2 \leq 1\}$. Then
$A = [-1,1]^r \times D^{(n - r)/2}$. Show like above that $\operatorname{vol}([-1,1]) = 2$ and $\operatorname{vol}(D) = \pi$. Then we got
$$\operatorname{vol}(A) = \operatorname{vol}([-1,1]^r \times D^{(n - r)/2}) = \operatorname{vol}([-1,1])^r \operatorname{vol}(D)^{(n - r)/2} = 2^r \pi^{(n - r)/2}.$$
This prove makes use of the fact you can write $A$ as $A = [-1,1]^r \times D^{(n - r)/2}$ and the relation between the one-dimensional Lebesgue measure and the multi-dimensional Lebesgue measure.
Hope it helps :)
