(Too long for a comment)
For the record:
$$\mathrm{exp\_abs}(3)=\frac{4}{3 \pi^2}\,I_3=0.3671989447$$
where
$$I_3=\int_{-1}^1\int_{-1}^1 \int_{-1}^1 |x+y+z|\sqrt{1-x^2}\sqrt{1-y^2}\sqrt{1-z^2}\,dx\,dy\,dz=2.718081241$$
(I couldn't express by known constants)
UPDATE:
The exact value of $\mathrm{exp\_abs}(3)$ can be deduced from the following (amazing!) result: if $X,Y,Z$ are independent and identically uniformly distributed on $S^3$ (the unit sphere in $\mathbb{R}^4$), then \begin{align*} \mathbb{E}|X+Y+Z|=W_3(1,1) &=\frac{476}{525}A+\frac{52}{7\,\pi^2}\frac{1}{A}\\ \end{align*} with $A:=\frac{3}{16}\frac{2^{1/3}}{\pi^4}\Gamma(\frac{1}{3})^6$. ( See https://scholarship.claremont.edu/jhm/vol6/iss1/7/ (on page 100). Timothy Budd pointed to this paper in the related post.)
If $X=(X_1,X_2,X_3,X_4)$ is uniform on $S^3$, and $U=(U_1,U_2)$ is uniform on $D^,$$D^2,$ (the unit disk in $\mathbb{R}^2$), the distributions of $X_1$ resp. $U_1$ coincide, and the same will then hold for the first coordinate of i.i.d. sums.
Now, for any rotationally symmetric random vector $V=(V_1,V_2,...,V_k)$ with finite expectation of $|V|:=\sqrt{V_1^2+V_2^2+\ldots +V_k^2}$ it holds that $$\mathbb{E}(|V|)=\frac{\Gamma(\tfrac{1}{2})\Gamma(\tfrac{k+1}{2})}{\Gamma(\tfrac{k}{2})}\mathbb{E}(|V_1|)$$
Thus if $U,V,W$ are uniform on $D^2$, and $X,Y,Z$ are uniform on $S^3$ we have \begin{align*}\mathbb{E}(|U+V+W|)=\frac{\pi}{2} \mathbb{E}(|(U+V+W)_1|\\ \mathbb{E}(|X+Y+Z|)=\frac{3\pi}{4} \mathbb{E}(|(X+Y+Z)_1|\end{align*} and therefore \begin{align*} \mathbb{E}(|U+V+W|)=\frac{2}{3}\,\mathbb{E}(|X+Y+Z|)=\frac{2}{3}W_3(1,1) \end{align*} and $$\mathrm{exp\_abs}(3)=\frac{2}{9}\,W_3(1,1)\;\;.$$ (And $I_3=\frac{\pi^2}{6}\, W_3(1,1)$.)