Find all $4$-tuples $(a,b,c,d)$ of natural numbers with $a \le b \le c$ and $a!+b!+c!=3^d$
1. We start with the equation \(a! + b! + c! = 3^d\) and the condition \(a \le b \le c\).
2. First, consider the case \(a = 1\). Then the equation becomes:
   \[
   1 + b! + c! = 3^d
   \]
   Simplifying, we get:
   \[
   b! + c! = 3^d - 1
   \]
3. Since \(3^d - 1\) is even, \(b!\) and \(c!\) must be such that their sum is even. This implies \(b\) and \(c\) must be at least 2 because \(1!\) is odd.
4. Consider \(b = 1\):
   \[
   1 + 1 + c! = 3^d \implies 2 + c! = 3^d
   \]
   For \(c \ge 3\), \(c!\) is divisible by 3, but \(3^d - 2\) is not divisible by 3. Hence, \(c\) cannot be 3 or greater. Therefore, \(c = 1\) or \(c = 2\).
   - If \(c = 1\):
     \[
     1 + 1 + 1 = 3^d \implies 3 = 3^d \implies d = 1
     \]
     This gives the solution \((a, b, c, d) = (1, 1, 1, 1)\).
   - If \(c = 2\):
     \[
     1 + 1 + 2 = 3^d \implies 4 = 3^d
     \]
     This is a contradiction since 4 is not a power of 3.

5. Consider \(b = 2\):
   \[
   1 + 2! + c! = 3^d \implies 1 + 2 + c! = 3^d \implies 3 + c! = 3^d
   \]
   Simplifying, we get:
   \[
   c! = 3^d - 3
   \]
   - If \(c = 2\):
     \[
     2! = 2 \implies 2 = 3^d - 3 \implies 3^d = 5
     \]
     This is a contradiction since 5 is not a power of 3.
   - If \(c = 3\):
     \[
     3! = 6 \implies 6 = 3^d - 3 \implies 3^d = 9 \implies d = 2
     \]
     This gives the solution \((a, b, c, d) = (1, 2, 3, 2)\).
   - If \(c = 4\):
     \[
     4! = 24 \implies 24 = 3^d - 3 \implies 3^d = 27 \implies d = 3
     \]
     This gives the solution \((a, b, c, d) = (1, 2, 4, 3)\).
   - If \(c = 5\):
     \[
     5! = 120 \implies 120 = 3^d - 3 \implies 3^d = 123
     \]
     This is a contradiction since 123 is not a power of 3.

6. For \(a > 1\), \(a! + b! + c!\) will be even, but \(3^d\) is odd for all \(d\). Hence, there are no solutions for \(a > 1\).

The final answer is \(\boxed{(1, 1, 1, 1), (1, 2, 3, 2), (1, 2, 4, 3)}\).