Edit: Time to add details, I think. The OP has long since worked it out but hasn't taken the invitation to write up a more complete solution, so I shall, in the interest of having a full answer to the question.
A pivot is a function of the data and the statistic whose distribution doesn't depend on the value of the statistic.
So consider:
(1) what would the distribution of a statistic consisting of the sum of the observations ($T=\sum_i x_i$) be?
A sum of $n$ i.i.d. $\text{gamma}(\alpha,\theta)$ random variables has the $\text{gamma}(n\alpha,\theta)$ distribution (for the shape-rate form of the gamma).
Here $n=6$ and $\alpha=2$, so the sum, $T$ has a $\text{gamma}(12,\theta)$ distribution.
(2) Note that the distribution in (1) does depend on $\theta$ and the form of the statistic doesn't. You need to modify the statistic ($Q=f(T,\theta)$) in such a way that both of those change. (This part is trivial!)
Let $Q=T/\theta$. Then $Q\sim \text{gamma}(12,1)$.
$Q$ satisfies the conditions required for a pivotal quantity.
(3) Once you have a pivotal quantity (i.e. $Q$), write down an interval for the pivotal quantity (in the form of a pair of inequalities, $a< Q< b$) with the given coverage. Since the distribution doesn't depend on the parameter, this interval is always the same (at a given sample size) no matter what the value of $\theta$.
One such interval is $(a,b)$, where $P(a<Q<b)=0.95$, when $a$ is the 0.025 point of the $\text{gamma}(12,1)$ distribution and $b$ is the 0.975 point.
(4) Now write the interval involving the pivotal quantity back in terms of the data and $\theta$. Back out an interval for the parameter, for which the corresponding probability statement must still hold (keeping in mind that the random quantity is not $\theta$ but the interval).
$P(a<T/\theta<b)=0.95$ implies $P(1/b < \theta/T < 1/a)=0.95$, so $P(T/b < \theta < T/a)=0.95$. Therefore $(T/b,T/a)$ is a 95% interval for $\theta$.
Our observed total, $t = 4.91$. The 0.025 point of a gamma(12,1) is 6.2006 and the 0.975 point is 19.682. Hence a 95% interval for $\theta$ is (4.91/19.682,4.91/6.200)
= $(0.249, 0.792)$.
First you need to find the distribution of the sufficient statistic $T=\max |X_i|$. You've already seen that $|X_i|=U\sim\mathrm{U}(0,\theta)$. To find the distribution of the maximum of $n$ observations, $T=U_{(n)}$, it's easiest to consider the cumulative distribution function:
$$\begin{align}
F_T(t) &= \Pr (U_{(n)} < t) = \Pr(U_1 <t, U_2 <t, \ldots, U_n <t)\\
&=F_U(t)^n= \left(\frac{t}{\theta}\right)^n
\end{align}$$
Differentiating with respect to $t$ gives the density
$$f_T(t)= \frac{nt^{n-1}}{\theta^n}$$
You can now calculate the density function of your proposed pivot $Q = \frac{T}{\theta}$ to confirm it's free of $\theta$:
$$f_Q(q)=f_T(\theta q)\cdot\left|\frac{\mathrm{d}t}{\mathrm{d}q}\right|=\frac{n(\theta q)^{n-1}}{\theta^n}\cdot\theta=nq^{n-1}$$
Best Answer
An outline of one way to approach it.
Step 1: show that $|X_i|/\theta$~Exponential(1)
Step 2: Hence explain how $\frac{2}{\theta}|X_i|\sim\chi^2_2$
Step 3: Hence give the distribution of $\frac{2}{\theta}\sum_i|X_i|$