I got to thinking I may have made it a little too hard because of the use of a pair of identities that may be stumbling blocks. So, my solution is below.

Solution: Plotting the successive differences listed in my previous post, one finds they form a parabola. The numerators are formed by the quadratic -(12 n^2 - 66n + 139). Thus we have the relation
f(n) - f(n-1) = -(12 n^2 - 66n + 139)/108

Using a telescoping sum, we can then compute f(n) where n starts at 0 as follows:
f(n) = f(n) - f(0)
= [f(n) - f(n-1)] + [f(n-1) - f(n-2)] + ... + [f(1) - f(0)]
= sum_k=0^(n-1) of f(n-k) - f((n-k)-1)
= -1/108 * sum_k=0^(n-1) of (12 (n-k)^2 - 66 (n-k) + 139)
= -1/108 * sum_k=0^(n-1) of (12 n^2 - 24 nk + 12 k^2 - 66n + 66k + 139)
= -1/108 * [[12 n^2 - 66n + 139] * n
- (24*n - 66) * sum_k=0^(n-1) of k
+ 12 * sum_k=0^(n-1) of k^2]
= -1/108 * [12 n^3 - 66 n^2 + 139 n - (24n - 66) * n * (n-1) / 2 + 12 * (n-1) * n * (2n-1) / 6]
= ... = -n^3 / 27 + n^2 / 4 + -n
= (-n/3)^3 + (-n/2)^2 + (-n/1)

The sums disappeared using the formulas sum_k=0^(n-1) of k = n*(n-1)/2 and sum_k=0^(n-1) of k^2 = (n-1)*n*(2n-1)/6. These are well-known identities and are special cases of Faulhaber's formula. These two can be proven inductively pretty easily (and IIRC I did so on this very forum a while back). The first has a cute geometric proof if you think in terms of triangles (indeed the numbers on the right side are called triangular numbers). The ... steps are just mindless simplification.