Is the nth-power-sum graph connected?
By going from $x$ to $a^n-x$ to $b^n-(a^n-x)$ we can, in two steps, add or subtract any difference of $n$th powers. In four steps, we can add or subtract any difference of differences (a second-order difference). And, repeating this idea, in $2^{n-1}$ steps, we can add or subtract any $(n-1)$th order difference. But the $(n-1)$th order differences of a sequence of $n$th powers are an arithmetic progression, so in this many steps we can reach from any initial $x$ the value $x$ mod $m$, where $m$ is the modulus of the arithmetic progression. That is, $\mathcal{G}_n$ has at most finitely many connected components.
To show that there is only one component, we need to show how to go from any value modulo $m$ to any other value modulo $m$. But this can be done, again, by going from $x$ to $a^n-x$ to $b^n-(a^n-x)$, where $a=m$ and $b=m+1$.
This shows, more strongly, that $\mathcal{G}_n$ has bounded diameter. I don't know how close to optimal is the exponential bound given by this argument.
Yes, the graph $\mathcal{G}_n$ is connected, and its diameter is at most $n2^n$. To see this, write $s$ for $n2^{n-1}$, and fix any two vertices $a,b\in\mathbb{N}^*$. By Wright's solution of Waring's problem with proportionality conditions, for a large parameter $c\in\mathbb{N}$, we can find $x_1,\dots x_s\in\mathbb{N}^*$ and $y_1,\dots y_s\in\mathbb{N}^*$ such that $$ x_1^n+\dots+x_s^n=c+a\qquad\text{and}\qquad y_1^n+\dots+y_s^n=c+b,$$ moreover the summands are asymptotically (as $c$ tends to infinity) $$ x_1^n,y_s^n\sim\frac{c}{2s-1}\qquad\text{and}\qquad x_2^n,\dots,x_s^n,y_1^n,\dots,y_{s-1}^n\sim\frac{2c}{2s-1}.$$ As a consequence, the following walk of length $2s$ connects $a$ and $b$: $$ a\ \rightarrow\ x_1^n-a\ \rightarrow\ -x_1^n+y_1^n+a\ \rightarrow\ x_1^n-y_1^n+x_2^n-a\ \rightarrow\ \cdots\ \rightarrow\ b.$$ Note that the edge sums in this walk are $x_1^n,y_1^n,\dots,x_s^n,y_s^n$, while each intermediate vertex is asymptotically $c/(2s-1)$, hence a positive integer for $c$ sufficiently large. The proof is complete.