Squaring both sides of the original equation, we get \(a^2 = 4+\sqrt{5+a}.\)

Subtracting 4 from both sides and squaring again gives \((a^2-4)^2 = 5+a.\)

Expanding this out and subtracting 5 + a from both sides, we have \(a^4-8a^2-a+11 = 0.\)

Similar manipulations on the other equations give \(\begin{align*} b^4-8b^2-b+11 &= 0, \\ c^4-8c^2+c+11 &= 0, \\ d^4-8d^2+d+11 &= 0. \end{align*}\)

(Note the signs carefully.)

Let \(f(x)=x^4-8x^2-x+11\). Then a and b are roots of f(x), while c and d are roots of \(x^4-8x^2+x+11\), which is f(-x). It follows that -c and -d are roots of f(x). Since -c and -d are negative and hence distinct from a and b, we have four roots of f(x). Since f(x) has degree four, we know we have found all the roots.

By Vieta's formulas, the constant term of f(x) is the product of the roots. Thus,

\(\begin{align} abcd = (a)(b)(-c)(-d) = \boxed{11}. \end{align}\)