The falling factorial is a product of a-0, a-1, a-2,...,a-(n-2).  The median of 0, 1, 2,...,n-2 is (n-2)/2, and it is a median of n-1 numbers.  The approximation comes by replacing each of the n-1 numbers subtracted off by their median.

Replace n with n+1 to get  a(a-1)...(a-n+1) ~ (a - (n-1)/2)^n.

Assume that a > n-1,take the natural logarithm, and divide by n to get

[ ln(a) + ln(a-1) + ... + ln(a-n+1) ] / n  ~  ln(a - (n-1)/2).

For n odd, this just says that the arithmetic mean is approximated by the median, while for n even, it says that the arithmetic mean is approximated by something close to the median. Reversing these steps provides a simple derivation of your approximation.

Another way to view this that avoids taking logarithms is simply to observe that the geometric mean [a(a-1)...(a-n+1)]^[1/n] is approximated by the median a - (n-1)/2.

Establishing the error in this approximation would be a logical next step. Intuitively, the approximation should improve as the argument a get larger.

The approximation is valid for large a, in fact for a >> n^(3/2), and the relative error behaves like -n^3/(24a^2), as can be estimated byTaylor-expanding the error in the logarithm for large a.