By the Trivial Inequality, $ (x - y)^2 \ge 0 $ for all real numbers $ x $ and $ y $. We can re-arrange this as \[x^2 + y^2 \ge 2xy.\]Equality occurs if and only if $ x = y $. (This looks like AM-GM, but we need to establish it for all real numbers, not just nonnegative numbers.) Setting $ x = a^2 $ and $ y = b^2, $ we get \[a^4 + b^4 \ge 2a^2 b^2.\]Setting $ x = c^2 $ and $ y = d^2, $ we get \[c^4 + d^4 \ge 2c^2 d^2.\]Setting $ x = ab $ and $ y = cd, $ we get \[a^2 b^2 + c^2 d^2 \ge 2abcd.\]Therefore \[a^4 + b^4 + c^4 + d^4 \ge 2a^2 b^2 + 2c^2 d^2 = 2(a^2 b^2 + c^2 d^2) \ge 4abcd.\]Since $ a^4 + b^4 + c^4 + d^4 = 48 $ and $ 4abcd = 48, $ all the inequalities above become equalities. The only way this can occur is if $ a^2 = b^2, $ $ c^2 = d^2, $ and $ ab = cd $. From the equations $ a^2 = b^2 $ and $ c^2 = d^2, $ $ |a| = |b| $ and $ |c| = |d| $. From the equation $ ab = cd, $ $ |ab| = |cd|, $ so $ |a|^2 = |c|^2, $ which implies $ |a| = |c| $. Therefore, \[|a| = |b| = |c| = |d|.\]Since $ abcd = 12, $ \[|a| = |b| = |c| = |d| = \sqrt[4]{12}.\]There are 2 ways to choose the sign of $ a, $ 2 ways to choose the sign of $ b, $ and 2 ways to choose the sign of $ c $. Then there is only 1 way to choose sign of $ d $ so that $ abcd = 12 $. (And if $ |a| = |b| = |c| = |d| = \sqrt[4]{12}, $ then $ a^4 + b^4 + c^4 + d^4 = 48 $.) Hence, there are a total of $ 2 \cdot 2 \cdot 2 = \boxed{8} $ solutions.