Simplifying the objective function: Since $p>0$ is a fixed quantity in this minimisation problem, the minimum point of $G_p$ will be the same as the minimum point of the following (simpler) objective function:
$$H_p(c) \equiv \begin{cases}
\sum_{i=1}^n |x_i - c|^p & & \text{if } p < \infty, \\[6pt]
\max_i |x_i - c| & & \text{if } p = \infty. \\[6pt]
\end{cases}$$
All we have done here is to remove the positive power $1/p$ from the function in the case where $0<p < \infty$. This is a monotonically increasing transformation of $G_p$, so it does not change the minimising point, but it makes the calculus for minimising the function simpler. As stated in the question, the absolute values can make the minimisation look more difficult. To deal with this we can write the absolute value in terms of the sign function as $|r| = r \cdot \text{sgn}(r) = -r \cdot \text{sgn}(-r)$, which gives:
$$|r|^p = r^p \cdot \text{sgn}^p (r) = \begin{cases}
r^p \cdot \text{sgn} (r) & & \text{if } p \text{ is odd}, \\[6pt]
r^p \cdot \mathbb{I} (r \neq 0) & & \text{if } p \text{ is even}, \\[6pt]
\end{cases}$$
Since $p>0$ we can re-write our objective function in its simplest form as:
$$H_p(c) \equiv \begin{cases}
\sum_{i=1}^n \text{sgn}(c-x_i) (c-x_i)^p & & \text{if } p \text{ is odd}, \\[6pt]
\sum_{i=1}^n (c-x_i)^p & & \text{if } p \text{ is even}, \\[6pt]
\max (|c - x_{(1)}|, |c - x_{(n)}|) & & \text{if } p = \infty. \\[6pt]
\end{cases}$$
As specified in the question, we will deal with the case where $p$ is an integer, so the above form exhausts all the possibilities here. Note that it is possible to extend to non-integer values of $p$, but we will not pursue that line of inquiry here.
Finding the measures of central location: We can find the measures of central location using ordinary calculus techniques, though we need to deal with some functions that have sharp points. The first derivative of the objective function is:
$$H'(c) = \begin{cases}
p \sum_{i=1}^n \text{sgn}(c-x_i) (c-x_i)^{p-1} & & \text{if } p \text{ is odd}, \\[6pt]
p \sum_{i=1}^n (c-x_i)^{p-1} & & \text{if } p \text{ is even}, \\[6pt]
\text{sgn}(c-\text{mid}(\mathbf{x})) & & \text{if } p = \infty, \\[6pt]
\end{cases} \quad \quad \quad$$
where $\text{mid}(\mathbf{x}) \equiv \tfrac{1}{2}(x_{(1)}+x_{(n)})$. The second derivative is:
$$\quad \quad \quad \quad H''(c) = \begin{cases}
2p \sum_{i=1}^n \delta (c-x_i) & & \text{if } p = 1, \\[6pt]
p(p-1) \sum_{i=1}^n \text{sgn}(c-x_i) (c-x_i)^{p-2} & & \text{if } p \neq 1 \text{ and } p \text{ is odd}, \\[6pt]
p(p-1) \sum_{i=1}^n (c-x_i)^{p-2} & & \text{if } p \text{ is even}, \\[6pt]
2 \delta (c-\text{mid}(\mathbf{x})) & & \text{if } p = \infty. \\[6pt]
\end{cases}$$
Since all of these values are either postive or at least non-negative, we can see that this function is strictly convex for $1 < p < \infty$ and weakly convex for $p = 1$. Setting $H'(\hat{c}) = 0$ gives the critical point which minimises the function, and therefore gives a measure of central location. Some particular solutions that are common measures of central location are:
$$\begin{aligned}
p=1 & & & & & \hat{c} = \text{median}(\mathbf{x}), \\[6pt]
p=2 & & & & & \hat{c} = \text{mean}(\mathbf{x}), \\[6pt]
\vdots \quad & & & & & \quad \quad \vdots \\[6pt]
p=\infty & & & & & \hat{c} = \text{mid}(\mathbf{x}). \\[6pt]
\end{aligned}$$
(Note that in the case where $p=1$ the minimising value is not necessarily unique --- if there are an even number of data points then any central point between the two middle values is a valid minimising point.) As can be seen, the media is the measure of central location that comes from penalising large deviations the least, and the mid-point is the measure that comes from penalising large deviations the most.
