Could someone tell me how large this number is?
Your function lies at $f_{\omega+4}$ in the fast-growing hierarchy; that is, your $f(n,n,n,n,n)$ is roughly $f_{\omega+4}(n)$.
First, let us set $g(n) = n \uparrow^{n+3} (n+1)$. Note that
$$2 \uparrow^{n-1}(n+1) < f_\omega (n) < 2 \uparrow^{n-1}(2n)$$
so for $n \ge 2$,
$$f_{\omega}(n) = f_n(n) < 2 \uparrow^{n-1}(2n) < g(n) < 2 \uparrow^{n+3}(n+5) < f_\omega(n+4)$$
Now we have $$f(a,0,0,0,0) = f(g(a),g(a),g(a),g(a),-1) = g(g(a))$$
Further, $$f(a,1,0,0,0) = f(g(a),0,0,0,0) = g(g(g(a))) = g^3(a)$$
and by induction we have $$f(a,b,0,0,0) = g^{b+2}(a)$$
In particular we have $f(a,a,0,0,0) = g^{a+2}(a)$, and
$$f_{\omega+1}(a) = f_{\omega}^a(a) < g^a(a) < g^{a+2}(a)$$
The other direction is a little tricker; however we can use the fact that $$f_{\omega}(a+1) > f_{\omega}(a) + 5$$ when $a \ge 2$, so setting $u(a) = f_{\omega}(a)$ and $v(a) = f_{\omega}(a+4)$, we have
$$u^n(a+5) = u^{n-1}(u(a+5)) = u^{n-1}(v(a+1)) > u^{n-1}(v(a)+5) = u^{n-2}(u(v(a)+5)) = u^{n-2}(v(v(a)+1)) > u^{n-2}(v(v(a)+5)) = \ldots > v^n(a)+5$$
Thus $$f_{\omega+1}(a+3) = u^{a+3}(a+3) = u^{a+2}(u(a+3)) > u^{a+2}(a+5) > v^{a+2}(a) > g^{a+2}(a)$$
and we have proven $$f_{\omega+1}(a) < f(a,a,0,0,0) < f_{\omega+1}(a+3)$$
The rest of the letters follow a similar pattern. Define $h(a) = g^a(a)$. Then we have
$$f(a,a,1,0,0) = f(g^a(a), 0, 1,0,0) = f(h(a),0,1,0,0) = f(g(h(a)),g(h(a)),0,0,0) = f(h(g(h(a))),0,0,0,0) = g(g(h(g(h(a)))))$$
$$f(a,a,2,0,0) = f(h(a),0,2,0,0) = f(g(h(a)),g(h(a)),1,0,0) = g(g(h(g(h(g(h(a))))))) = g(gh)^3(a)$$
and by induction again,
$$f(a,a,c,0,0) = g(gh)^{c+1}(a)$$
Now both $gh(a) = g^{a+1}(a)$ and $ggh(a) = g^{a+2}(a)$ lie between $f_{\omega+1}(a)$ and $f_{\omega+1}(a+3)$, so $f(a,a,c,0,0) =g(gh)^{c+1}(a)$ is greater than $f_{\omega+1}^{c+1}(a)$, and by the same argument as the $u,v$ argument above we can show that $f(a,a,c,0,0)$ is less than $f_{\omega+1}^{c+1}(a+4)$.
So $$f_{\omega+2}(a) = f_{\omega+1}^a(a) < f_{\omega+1}^{a+1}(a) < f(a,a,a,0,0) < f_{\omega+1}^{a+1}(a+4) < f_{\omega+1}^{a+1}(f_{\omega+1}(a+2)) = f_{\omega+1}^{a+2}(a+2) = f_{\omega+2}(a+2)$$
Using the same reasoning, we can show that $$f_{\omega+3}(a) < f(a,a,a,a,0) < f_{\omega+3}(a+2)$$ and $$f_{\omega+4}(a) < f(a,a,a,a,a) < f_{\omega+4}(a+2)$$
In terms of Conway chained arrow notation, $f(a,a,a,a,a)$ will be approximately $a \rightarrow a \rightarrow a+1 \rightarrow 5$.
In terms of BEAF, $f(a,a,a,a,a)$ will be approximately $\lbrace a,a+1,4,2 \rbrace$.
In BEAF notation, I have deduced that
$$f(a,0,0,0,0)\approx\{a,2,1,2\}$$
$$f(a,b,0,0,0)\approx\{a,b+1,2,2\}$$
$$f(a,b,1,0,0)\approx\underbrace{\{a,\{a,\{a,\{a,\{}_{\{a,b+1,2,2\}}\dots\}+1,2,2\}+1,2,2\}+1,2,2\}+1,2,2\}$$
$$f(a,b,c,0,0)\approx(c+1)\left\lbrace\tiny\underbrace{\{a,\{a,\{a,\{a,\{a,\{a,\{a,\{a,\{a,\{a,\{a,\{a,\{a,\{a,\{a,\{a,\{}_{\underbrace{\{a,\{a,\{a,\{a,\{a,\{a,\{a,\{a,\{a,\{a,\{a,\{a,\{a,\{a,\{a,\{a,\{}_{\underbrace{\{a,\{a,\{a,\{a,\{a,\{a,\{a,\{a,\{a,\{a,\{a,\{a,\{a,\{a,\{a,\{a,\{}_{\underbrace{\vdots}_{\{a,b+1,2,2\}}}\dots\}+1,2,2\}+1,2,2\}+1,2,2\}+1,2,2\}}\dots\}+1,2,2\}+1,2,2\}+1,2,2\}+1,2,2\}}\dots\}+1,2,2\}+1,2,2\}+1,2,2\}+1,2,2\}\right.$$
Unless someone can simplify this, BEAF isn't going to be enough, but it looks like a good start.
Maybe BEAF is a good place, but I'm going to have to understand more about it's nature for more terms in the array and larger numbers at the end.
:-(