You don't have to find the $a'$ s

Since $x_n$ is a linear combination of the sequences $1, n, n^2, \cdots, n^9,$ then it satisfies a linear recurrence whose characteristic polynomial is $(r-1)^{10}=\sum_{k=0}^{10} \binom{10}{k}(-1)^{10-k}r^{10-k}.$ Then we have that $x_n=\sum_{k=1}^{10}\binom{10}{k}(-1)^{11-k}x_{n-k} \quad\text{for all}\quad n\geq 11.$ For the details see Linear Recurrence Relations - With Repeated Roots . Then making $n=11,$ we obtain $x_{11}=\sum_{k=1}^{10}\binom{10}{k}(-1)^{11-k}x_{11-k}= \sum_{k=1}^{10}\binom{10}{k}(-1)^{11-k}e^{11-k}=e(e^{10}-(e-1)^{10})\approx$ $\approx 59264.2700\cdots$ Therefore, $\left\lfloor x_{11}\right\rfloor=\boxed{59264}.$