Suppose each $x_{n+1} \le \left(\sum_{i=1}^n x_i \right)^{-c}$ for some $c \in (0,1)$. How quickly can $\sum_{i=1}^n x_i $ grow?

The claim for the sum holds true: if $s_n=\sum_{k=1}^n x_k$ then $n\mapsto s_n^{c+1}/n$ is bounded.

Here's a fact that helps: for any sequence $n\mapsto a_n$ of real numbers, we have $$\limsup_{n\to\infty}(a_n/n)\leqslant\limsup_{n\to\infty}(a_{n+1}-a_n)$$ (this is a corollary of the Stolz–Cesàro theorem, but is also easy to prove directly).

From the premises, $s_n\leqslant s_{n+1}\leqslant s_n+s_n^{-c}$; with $a_n=s_n^{c+1}$, this can be rewritten as $$0\leqslant a_{n+1}-a_n\leqslant\psi(1/a_n),\qquad\psi(t):=\frac{(1+t)^{c+1}-1}{t}.$$ Observe that $\lim_{t\to 0^+}\psi(t)=c+1$ is finite, thus $\psi$ is bounded on $(0,a)$ for any $a>0$.

Hence, $n\mapsto a_{n+1}-a_n$ is bounded, and by the above, $n\mapsto a_n/n$ is bounded as well.

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The claim regarding $x_n$ may fail to hold. We can violate it on a subsequence like $x_{2^n}$. Specifically, take any "slowly convergent" $\sum_{k=0}^\infty a_k=1$, say with $a_k=1/[(k+1)(k+2)]$, and let $x_n=a_k$ if $n=2^k$ and $x_n=0$ otherwise (for $x_n\neq 0$, take $x_n=a_k$ if $n=2^k$ and $k>0$, and $x_n=2^{-n}a_0$ otherwise).