This can be done by hand (although I don't know about the qualifier "reasonably").
As done in the other answer (slightly modified), we can let $X = \frac{A-D}{2}$, which since $A$ and $D$ are both $\sim U(-1,1)$ has a triangular distribution centered at $0$ and extending from $-1$ to $1$. If we consider $|X|$, this has a CDF of $P(|X|<x) = 2x-x^2$ for $x \in (0,1)$. Since $|X|$ is positive, we have $P(X^2 < x) = P(|X|<\sqrt{x}) = 2\sqrt{x} - x$. Differentiating gives the PDF $f_{X^2}(x)=\frac{1}{\sqrt{x}} - 1$, for $x \in [0,1]$.
From symmetry, we can see from the joint distribution of $A$ and $D$ that the distribution of $X$ when $A+D>0$ is the same as the distribution of $X$ when $A+D<0$, and as stated in the post, these are equally likely, producing a factor of 1/2.
We can also define $Y=-BC$. If $Y < 0$, then clearly $X^2 > Y$. We have $P(Y<0)=\frac{1}{2}$, again by symmetry.
If $Y\ge 0$ (which occurs half the time), then we want to know if $X^2>|Y|=|B||C|$, which is the product of two independent rv $\sim U(0,1)$, so $|Y|$ has the PDF $f_{|Y|}(y) = -\log(y)$ for $ y \in (0,1)$.
see product distribution of two uniform distribution, what about 3 or more
Since $X^2$ and $|Y|$ are independent, their joint distribution is the product of their distributions, so we want to integrate:
$$\frac{1}{2}\left[ \frac{1}{2} + \frac{1}{2} \int_{y=0}^{y=1} \int_{x=y}^{x=1}\left(\frac{1}{\sqrt{x}} - 1\right)\left(-\log(y)\right)dx\, dy \right]$$
Where the first $\frac{1}{2}$ is from the factor for $A+D>0$, the next $\frac{1}{2}$ is for when $Y<0$, and the next $\frac{1}{2}$ is from the factor for $Y>0$.
For the inner integral:
$$\int_{x=y}^{x=1}\left(\frac{1}{\sqrt{x}} - 1\right) dx= (1+y-2\sqrt{y})$$
For the outer integral:
$$\int_{y=0}^{y=1} (1+y-2\sqrt{y})\left(-\log(y)\right)dy$$
Letting $u = \sqrt{y}$, so that $dy = 2u \,du$:
\begin{align}
\int_{u=0}^{u=1} (1+u^2-2u)\left(-2\log(u)\right)2u\, du
&= -4 \int_{u=0}^{u=1} (u-2u^2+u^3)\left(\log(u)\right) du \\
&= -4 \left(\int_{u=0}^{u=1} u\log(u) du - 2\int_{u=0}^{u=1} u^2\log(u) du + \int_{u=0}^{u=1} u^3\log(u) du \right) \\
&=4 \left( \frac{1}{4} - \frac{2}{9} + \frac{1}{16} \right) \\
&=1-\frac{8}{9}+\frac{1}{4} \\
&=\frac{13}{36} \\
\end{align}
For the integrals, I looked up in a table that for $m \ge 0$:
$$\int x^m \log(x)dx = \frac{x^{m+1}}{m+1} \left( \log(x) - \frac{1}{m+1} \right)$$
Putting this in the main equation gives the probability:
$$\frac{1}{2}\left[ \frac{1}{2} + \frac{1}{2} \frac{13}{36} \right] = \boxed{ \frac{49}{144} } \approx 0.3403$$
