Suppose we define a class of functions $f_n$ (with domain $\mathbb{N}^n$ without $0$) as follows: $$ f_n(x_1,x_2,\ldots,x_n)=\begin{cases} 1, & x_1=1\\ x_1, & x_2=1\\ \,\vdots & \,\vdots\\ x_{n-2}, & x_{n-1} = 1 \\ x_{n-1}, & x_n = 1\\ f_n(x_1-1,x_2,\ldots,x_n) + \ldots + f_n(x_1,x_2,\ldots,x_n-1), &\text{otherwise} \end{cases} $$ For clarity, the otherwise case returns the value $$ f_n(x_1-1,x_2,\ldots,x_n) + f_n(x_1,x_2-1,\ldots,x_n) + \ldots + f_n(x_1,x_2,\ldots,x_n-1). $$
We can see that $f_1(x) = 1$ for all $x$, and from the similarity to binomial coefficients we get that $$f_2(x,y) = {{x+y-1}\choose{x-1}}. $$ It may be helpful to observe that $f_n$ has a similar recursive definition to that of the multinomial coefficient.
The question is:
Is there a closed form formula of $f_n(x_1,x_2,\ldots,x_n)$ for any given $n$, and what is it given that one exists?
Clarification: Evaluate cases in order from top to bottom. For example, if both $x_n = 1$ and $x_{n+k} = 1$, the case for $x_n$ takes precedent.