Not always, no. They measure different things and fail in different ways.
Take the dataset [7, 8, 9], with a median and mean of 8. Adding a 100 to the set results in a median of 8.5 and a mean of 31, so the mean moves much farther. This is probably the effect you're thinking of: the mean can take extreme values into account "too much".
But I can also make the median move more. Take the dataset [0, 50, 100]. The median and mean are both 50. If I add [100, 100] to it so it becomes [0, 50, 100, 100, 100], the mean moves to only 70, but the median moves all the way up to 100! There was a "gap" in the numerical sequence that the median could jump over, but the mean couldn't.
Here's a different way of moving the median further. Take the dataset [1, 1, 1, 1, 2, 3, 4, 5, 5, 5, 5]. Bathtub-shaped data. As I add fives to the set, the mean goes 3, 3.17, 3.3. But the median goes 3, 3.5, 4! Medians move past thin spots in distributions very quickly.
Mean is sensitive to distant outliers; median is sensitive to unevenly distributed data and numerical gaps.
To come back on topic, while I don't have a reference for the age-at-death distribution, I think it's bathtub-shaped. Hence, the median might be more sensitive to extra values at the top than the mean would be.
While those are interesting counterexamples, they don't come close to modeling a realistic "age-at-death" distribution. There isn't any such distribution whose median would be significantly skewed by a tiny minority of centenarian outliers. These distributions are basically unimodal with the exception of a some degree of infant mortality, and the average (median or mean) is not located at a thin spot in the distribution (quite the opposite) [1].
Could you elaborate? Isn't the median always _less_ sensitive to outliers compared to the mean?