Is there such a thing as backwards sigma?

Sure:

$$\sum_{i=0}^{9}(10-i)=10+9+8+\ldots+1$$

The problem with your example is that addition is commutative, so it is not really useful to have a distinction for a sum from $1$ to $10$ or from $10$ to $1$.

However, your question makes sense in a noncommutative setting. Suppose for instance you have 10 matrices $A_1$, ..., $A_{10}$. Since the product of matrices is not in general commutative, the product $A_1 \dotsm A_{10}$ is in general different from $A_{10} \dotsm A_1$. In this case, you may consider writing $\prod_{i=1}^{10}A_i$ in the first case and $\prod_{i=10}^{1}A_i$ in the second case, but this is probably not the most satisfying solution.

A better solution is to consider a totally ordered finite set $(I, \leqslant)$ and to write $\prod_{i \in I}A_i$. The first case of my example can now be obtained by considering the set $\{1, \ldots, 10\}$ ordered by $1 \leqslant 2 \dotsm \leqslant 10$ and the second one by considering the set $\{1, \ldots, 10\}$ ordered by $10 \leqslant 9 \dotsm \leqslant 1$.

Try $$ \sum_{i=1}^{10} (11 - i) $$ There is no reason to introduce another symbol when a simple subtraction can do the work.