In a general sense, the answer is no: we certainly get nothing new at the level of absolute convergence, and we get little concrete new even when we look at ordinal-length sums.
First, let's look at the situation where all the reals we're summing are nonnegative (so convergence and absolute convergence coincide, and we don't have to worry about order in the usual setting). If $(a_i)_{i\in I}$ is a family of nonnegative reals and $I$ is uncountable, one of the following must occur: either all but countably many $a_i$s are zero, or there is some $\epsilon>0$ such that $$\{i: a_i>\epsilon\}$$ is uncountable.
- Why? Well, for each positive rational $q$, let $X_q=\{i: a_i>q\}$. Then we have that $$I=[\bigcup_{q\in\mathbb{Q}_{>0}} X_q]\cup\{i: a_i=0\}.$$ This expresses the uncountable set $I$ as a union of countably many sets (since there are only countably many positive rationals), and so one of those sets must be uncountable; if it's $\{i:a_i=0\}$ then we're in the first case, and if it's one of the $X_q$ then just take $\epsilon$ to be such a $q$.
Now it's easy to argue that - if each of the $a_i$s is nonnegative - we need to set $\sum_{i\in I}a_i=+\infty$ (or "diverges" if you prefer) unless all but countably many of the $a_i$s are zero: if uncountably many of the $a_i$s are nonzero, then we get infinitely many $a_i$s which are above some fixed positive real, and that goes to infinity.
So when everything has the same sign, uncountable sums reduce to classical countable sums. What about when different signs are allowed?
The usual definition of sum over an arbitrary index set looks at the set of all finite partial sums. However, this doesn't take into account order, and so gives the answer "undefined" for (the unordered versions of) sums which conditionally converge. We might reasonably hope that we could bring order into the situation somehow to get a more interesting picture.
One natural candidate for this is to use arbitrary ordinals (this bit quickly gets technical; I hope I've made it somewhat understandable). We can define (in the sense of "give values to or set to 'undefined'") "sums of length $\gamma$" for an arbitrary ordinal $\gamma$ by recursion on $\gamma$:
Base clause: The empty sum is zero.
Successor clause: If $(a_\eta)_{\eta<\theta+1}$ is an $(\theta+1)$-length sequence of reals and we've already analyzed all $\theta$-length sums, then $$\sum_{\eta<\theta+1}a_\eta=(\sum_{\eta<\theta}a_\eta)+a_{\theta}$$ if $\sum_{\eta<\theta}a_\eta$ exists, and is undefined otherwise.
Limit clause: If $\lambda$ is a limit ordinal, $(a_\eta)_{\eta<\lambda}$ is a $\lambda$-length sequence of reals, and we've already analyzed every $\theta$-length sum for every $\theta<\lambda$, we set $\sum_{\eta<\lambda}a_\eta$ to be $L$ if $(i)$ for all sufficiently large $\theta<\lambda$ the sum $\sum_{\eta<\theta}a_\eta$ is defined and $(ii)$ the limit of the values of the partial sums $\sum_{\eta<\theta}a_\eta$ for $\theta<\lambda$ exists and equals $L$ (this can be defined precisely via topology; I'm ignoring that issue for now), and undefined otherwise.
This looks attractive for a bit, and is not fully trivial. However, it still isn't really that different from the usual $\mathbb{N}$-length sum. For example, general results from descriptive set theory indicate that - looking at the least uncountable ordinal $\omega_1$ - if $(a_\eta)_{\eta<\omega_1}$ is a "reasonably definable" $\omega_1$-sequence of reals and $\sum_{\eta<\omega_1}a_\eta$ as defined above exists, then it in fact equals the values of "most" of the partial sums (specifically: there is a club of $\theta$ on which $\sum_{\eta<\theta}a_\eta$ is defined and equals $\sum_{\eta<\omega_1}a_\eta$). This is all going a bit advanced, but the point is that we're not getting a lot of new behavior in concrete cases. So this explains why even the above notion, which avoids the usual trivialities, doesn't really show up.
