To achieve this, we have first computed all sums of 2, 3 and 4 cubes less than 10^9. There are 440959 sums of 2 cubes, 100735175 sums of 3 cubes, and 966039273 sums of 4 cubes. The corresponding program is the following.

Then for each a such that 108 <= a <= 15000, we have checked that each x such that a^3 <= x < (a+1)^3, the integer x can be written as a sum of five cubes. For this, we have first looked for x-a^3 in the table of 4 cubes, in case of failure for x-(a-1)^3, until we found the smallest k such that x-(a-k)^3 is a sum of 4 cubes. (If the table of 4 cubes was too small, then we substracted the largest possible cube and looked in the table of 3 cubes.) Let us denote this smallest k by k(x). The checking program is called "check a0 a1 N" and checks all values of x such that a0^3 <= x < (a1+1)^3, assuming that tables of size N have been computed. For each value of a such that a0 <= a <= a1, the program outputs a list containing the number of x, a^3 <= x < (a+1)^3, such that k(x)=0, 1, ... For example, "check 1000 1001 10000000" outputs:
1000 [2312413,459836,183438,35685,5545,4757,1024,180,99,22,1,1] 10000000
1001 [2317306,588491,52874,37397,10504,1268,862,259,32,12,2] 10000000
which tells us that in 1000^3 <= x < 1001^3, there are 2312413 numbers for which k(x)=0, i.e. x-1000^3 is a sum of 4 cubes, 459836 numbers with k(x)=1, ..., and 1 number with k(x)=11. For a=1001, the largest value of k(x) is 10.

Finally, here is a Maple file containing the lists of outputs of the program for 108 <= a <= 15000.