Alpoge's comment gives a beautiful answer for prime bases. Of course this answers the original question for binary addition.

I can't see how to extend it to composite bases. For instance in base $6$, the addition $2+7 = 9 = 13_6$ has one carry, but the highest power of $6$ dividing $\binom{9}{2} = 36$ is $2$.

Here is an alternative argument. Work in base $d$. If $x = x_r d^r + \cdots + x_jd^j + \cdots x_1 d + x_0$ then say $x_j$ is in position $j$. Suppose we are adding $a^{(1)}, \ldots, a^{(n)}$ and that there is a carry of $r$ into position $j$. Consider the sum $r + a^{(1)}_j + \cdots + a^{(n)}_j$, computed in $\mathbb{N}_0$. We get a carry every time a partial sum $r + a^{(1)} + \cdots + a^{(m)}_j$ is $< ds$ (for some $s$) and the next addition makes it $\ge ds$ (but necessarily $< (d+1)s$). So if the total is $y \in \mathbb{N}_0$ then the number of carries is just $\lfloor y/d \rfloor$. Since addition in $\mathbb{N}_0$ is commutative and associative, this does not depend on the order of the numbers. Neither does the carry going into position $j+1$. So by induction, the total number of carries is independent of the order of the numbers.

Alpoge's comment gives a beautiful answer for prime bases. Of course this answers the original question for binary addition.

I can't see how to extend it to composite bases. For instance in base $6$, the addition $2+7 = 9 = 13_6$ has one carry, but the highest power of $6$ dividing $\binom{9}{2} = 36$ is $2$.

Here is an alternative argument. Work in base $d$. If $x = x_r d^r + \cdots + x_jd^j + \cdots x_1 d + x_0$ then say $x_j$ is in position $j$. Suppose we are adding $a^{(1)}, \ldots, a^{(n)}$ and that there is a carry of $r$ into position $j$. Consider the sum $r + a^{(1)}_j + \cdots + a^{(n)}_j$, computed in $\mathbb{N}_0$. We get a carry every time a partial sum $r + a^{(1)} + \cdots + a^{(m)}_j$ is $< ds$ (for some $s$) and the next addition makes it $\ge ds$ (but necessarily $< (d+1)s$). Since addition in $\mathbb{N}_0$ is commutative and associative, the number of such 'crossings' does not depend on the order of the numbers. Neither does the carry going into position $j+1$. So by induction, the total number of carries is independent of the order of the numbers.