Answer:
Although it does not resemble any distribution I know, it is possible to obtain a compact expression for the probability density. Denote $W=p_1+p_2$ and $Z=p_1+p_3$ then their joint density takes form:
$$f(w,z) = \int_{w+z-1}^{1} h(\xi,w-\xi,z-\xi) d\xi,$$
where $h(p_1,p_2,p_3) = \frac{p_1^{a_1-1} p_2^{a_2-1} p_3^{a_3-1}(1-p_1-p_2-p_3)^{a_4-1}}{B(a_1,a_2,a_3,a_4)}$ is the pdf of $Dir(a_1,a_2,a_3,a_4)$.
Solution:
(apologies for volume, comments are very welcome)
Denote $F(w,z)$ the cumulative density function of the distribution of $(p_1+p_2,p_1+p_3)$, where $p_1,p_2,p_3$ jointly with some $p_4$ are distributed as $Dir(a_1,a_2,a_3,a_4)$. By definition of CDF
$$ F(w,z) = P(\{p_1 + p_2 < w \}\cap\{p1+p3<z\}\cap\{p1+p2+p3 \leq 1\}) $$
where each expression in curly braces denotes a random event.
Now, we just have to unravel that complex event into a union of disjoint ones:
\begin{align}
&\begin{cases}
p_1 + p_2 < w \\
p1+p3<z \\
p1+p2+p3 \leq 1
\end{cases}
\iff\\
\iff
\begin{cases}
p_1 < w+z - 1 \\
p_2 \in [0, 1 - z) \\
p_3 \in [0, z - p_1)
\end{cases}
\text{ OR }
&\begin{cases}
p_1 < w+z - 1 \\
p_2 \in [1 - z, w - p_1) \\
p_3 \in [0, 1 - p_1)
\end{cases}
\text{ OR }
\begin{cases}
p_1 > w+z - 1 \\
p_2 \in [0, w - p_1) \\
p_3 \in [0, z - p_1)
\end{cases}
\end{align}
The cumulative density function is thus a sum of three components:
$$ F(w,z) = F_1(w,z) + F_2(w,z) + F_3(w,z), $$ where
\begin{align}
F_1(w,z) = &\int_0^{w+z-1}\int_0^{1-z} \int_0^{z-p_1} h(p_1,p_2,p_3) dp_3 dp_2 dp_1 \\
F_2(w,z) = & \int_0^{w+z-1}\int_{1-z}^{w-p_1} \int_0^{1-p_1-p_2} h(p_1,p_2,p_3) dp_3 dp_2 dp_1 \\
F_3(w,z) = & \int_{w+z-1}^{1}\int_{0}^{w-p_1} \int_0^{z-p_1} h(p_1,p_2,p_3) dp_3 dp_2 dp_1. \\
\end{align}
The simplest way ahead would be to obtain the probability density function $f(w,z) = \frac{\partial^2}{\partial z\partial w} F(w,z)$. Derivatives of components $F_1,F_2,F_3$ with respect to $w$ look like
\begin{align}
\frac{\partial}{\partial w}F_1(w,z) = &\int_0^{1-z} \int_0^{1-w} h(w+z-1,p_2,p_3) dp_3 dp_2 \\
\frac{\partial}{\partial w}F_2(w,z) = & \int_0^{w+z-1}\int_{1-z}^{1-z} \int_0^{1-p_1-p_2} h(p_1,p_2,p_3) dp_3 dp_2 dp_1 + \int_0^{w+z-1} \int_0^{1-w} h(p_1,w-p_1,p_3) dp_3 dp_1 \\
=& \int_0^{w+z-1} \int_0^{1-w} h(p_1,w-p_1,p_3) dp_3 dp_1 \\
\frac{\partial}{\partial w} F_3(w,z) = & - \int_{0}^{1-z} \int_0^{1-w} h(w+z-1,p_2,p_3) dp_3 dp_2 + \int_{w+z-1}^{1} \int_0^{z-p_1} h(p_1,w-p_1,p_3) dp_3 dp_1. \\
\end{align}
Note that $\frac{\partial}{\partial w}F_1(w,z)$ cancels out with the first term of the last expression. The final PDF takes form:
\begin{align}
f(w,z) =& \frac{\partial}{\partial z} \left(\int_0^{w+z-1} \int_0^{1-w} h(p_1,w-p_1,p_3) dp_3 dp_1 + \int_{w+z-1}^{1} \int_0^{z-p_1} h(p_1,w-p_1,p_3) dp_3 dp_1\right) =\\
=&\int_0^{1-w} h(w+z-1,1-z,p_3) dp_3 - \int_0^{1-w} h(w+z-1,1-z,p_3) dp_3 +\\ &+ \int_{w+z-1}^{1} h(p_1,w-p_1,z-p_1) dp_1 =\\
=& \int_{w+z-1}^{1} h(p_1,w-p_1,z-p_1) dp_1
\end{align}
and we stop here.
