Using the Pochhammer symbol notation, let $$S_0 = \sum_{n=0}^\infty \frac{(1/2)_n^3}{n!^3}\sum_{k=1}^n \frac{1}{2k-1} \qquad S_1 = \sum_{n=0}^\infty \frac{(1/2)_n^3}{n!^3}H_n$$
with $H_n$ the Harmonic number. OP's sum is $S_0$. I will show
$$\tag{0}S_0 = \frac{\Gamma \left(\frac{1}{4}\right)^4}{24 \pi ^2} \qquad S_1=\frac{(\pi -3 \log (2)) \Gamma \left(\frac{1}{4}\right)^4}{6 \pi ^3}$$
(I realized after posting this answer, that all details presented below were in fact very succinctly presented here).
First note that derivative of Pochhammer symbol can be expressed as digamma function: ${\left. {\frac{d}{{d\varepsilon }}} \right|_{\varepsilon = 0}}{(a + \varepsilon )_n} = {(a)_n}\left[ {\psi (a + n) - \psi (a)} \right]$. Consider $$\begin{aligned}&\quad {\left. {\frac{d}{{d\varepsilon }}} \right|_{\varepsilon = 0}}{_3F_2}\left( {\frac{1}{2} + \varepsilon ,\frac{1}{2} + \varepsilon ,\frac{1}{2};1 + \varepsilon ,1;1} \right)\\
&= \sum\limits_{n = 0}^\infty {\frac{{{(1/2)}_n^3}}{{n{!^3}}}\left[ {2\psi (n + \frac{1}{2}) - 2\psi (\frac{1}{2}) - \psi (1 + n) + \psi (1)} \right]} = 4S_0 - S_1
\end{aligned}$$
On other other hand, the above $_3F_2$ can be evaluated to be $$\frac{2^{-\varepsilon-\frac{1}{2}} \Gamma \left(\frac{1}{4}-\frac{\varepsilon}{2}\right) \Gamma (\varepsilon+1)}{\Gamma \left(\frac{3}{4}-\frac{\varepsilon}{2}\right) \Gamma \left(\frac{\varepsilon}{2}+\frac{3}{4}\right)^2}$$
calculating its derivative gives $$\tag{1}4S_0 - S_1 = \frac{\log (2) \Gamma \left(\frac{1}{4}\right)^4}{2 \pi ^3}$$
We need to find a second equation relating $S_0,S_1$, this is more difficult. I will state the following result, it will be proved at the end.
For $0<a<1,0<x\leq 1/2$, we have
$$\tag{*}{_2F_1}\left( {a,1 - a;1;x} \right){_2F_1}\left( {a,1 - a;1;1 - x} \right) = \frac{{\sin a\pi }}{\pi }\sum\limits_{n = 0}^\infty {{c_n}\left[ {{d_n} - \log (4x(1 - x))} \right]{{(4x(1 - x))}^n}} $$
where $${c_n} = \frac{{{{(1/2)}_n}{{(a)}_n}{{(1 - a)}_n}}}{{n{!^3}}}\qquad {d_n} = 3\psi (n + 1) - \psi (n + \frac{1}{2}) - \psi (n + a) - \psi (n + 1 - a)$$
Let $a=x=1/2$,
$$\begin{aligned}\pi {_2F_1}{\left( {\frac{1}{2},\frac{1}{2};1;\frac{1}{2}} \right)^2} &= \sum\limits_{n = 0}^\infty {\frac{{{(1/2)}_n^3}}{{n{!^3}}}\left[ {3\psi (n + 1) - 3\psi (n + \frac{1}{2})} \right]} \\
&= 3S_1 - 6S_1 +6\log 2 \sum\limits_{n = 0}^\infty {\frac{{{(1/2)}_n^3}}{{n{!^3}}}}
\end{aligned}$$
the $_3F_2$ and $_2F_1$ appear at both sides are easy (for example, an elliptic singular value for LHS, Dixon's theorem for RHS), so
$$\tag{2}3S_1 - 6S_0 = \frac{\Gamma \left(\frac{1}{4}\right)^4}{4 \pi ^2}-\frac{\Gamma \left(\frac{1}{4}\right)^4}{4 \pi ^3}(6\log 2)$$
Solving $(1), (2)$ gives $(0)$.
I briefly mention a proof of (*). For $a=1/2$, this was discovered by Watson in 1908, the following proof works essentially uniformly for all $0<a<1$.
An one-line proof: Verify both sides of $(*)$ satisfy the following 3rd order ODE:
$$(-4 a^2 x^2+4 a^2 x+4 a x^2-4 a x+6 x^2-6 x+1) y'(x)-2 (a-1) a (2 x-1) y(x)+3 (x-1) (2 x-1) x y''(x)+(x-1)^2 x^2 y^{(3)}(x)=0$$
and their first few terms of series expansion at $x=0$ agree.
A conceptual proof: $y_1(x) = {_2F_1}(a,1-a;1;x)$ satisfies the following ODE: $$(1-a) a y(x)+(2 x-1) y'(x)+(x-1) x y''(x)=0$$ which is invariant under $x\mapsto 1-x$, so $y_2(x) = y_1(1-x)$ is the other solution. This ODE is Fuchsian with a logarithmic singularity at $x=0$. On the other hand, $y_1^2 = {_3F_2}(a,1-a,1/2;1,1;4x(1-x))$. Denote $X=4x(1-x)$. $y_1^2, y_1y_2, y_2^2$ satisfies a Fuchsian 3rd order ODE in $X$. Theory of Fuchsian equation says, if $y_1(x)^2 = \sum c_n X^n$, then $$y_1(x)y_2(x) = A \log X\sum c_n X^n + B \sum d_nX^n$$
$$y_2(x)^2 = C \log^2 X \sum c_n X^n + D \log X \sum d_n X^n + E \sum e_n X^n$$
here $d_n$ (resp. $e_n$) can be explicitly given, via Frobenius method with integral exponent differences, to be $c_n$ times digamma (resp. trigamma) factors. This explains the form of $(*)$, making all these explicit is not difficult.
Best Answer
The convolution of the indicator functions extends up to the sum of their widths. That is, the support of the convolution $\circledast_k I_{[-a_k,a_k]}$ is $\left[-\sum_ka_k,\sum_ka_k\right]$. The convolution is a non-negative function, so its integral from $1$ to $\infty$ is positive iff its support extends beyond $1$, that is, if $\sum_ka_k\gt1$.