From your comment, it seems that you may not be aware that the outer product is equivalent to a sum in the exponent. Using $\exp(a)\exp(b)=\exp(a+b)$ and transforming the indices as suggested in a comment, you can rewrite your function as
$ f(x_1,x_2,\ldots,x_n) = C\exp\left\{\sum_{m=1}^n\left(-\frac{1}{2a^2}\left(x_m-\sum_{k=m-L}^{m-1}\alpha_{m-k}x_k\right)^2+y_m\sum_{l=0}^L\beta_{m-k}x_k\right)\right\}\;. $
You can see from this representation that the matrix of the quadratic form (whose determinant you need) is a band matrix with bandwidth $2L+1$. The diagonal entries are constant ($1+\sum_i\alpha_i^2$) up to $m=n-L$, but decrease for greater $m$ (because successive $\alpha_i^2$ go missing) down to $1$ for $m=n$. Because of this asymmetry, I doubt that you'll find a nice closed form for the determinant.