A polynomial $p(x)$ of degree $n\ge 2$ has exactly $n$ real roots, counted with multiplicity. We know that the coefficient of $x^n$ is $1$, all the roots are less than or equal to $1$, and $p(2)=3^n$. What values can $p(1)$ take?
1. **Restate the problem in terms of the roots:**
   Given a polynomial \( p(x) \) of degree \( n \) with exactly \( n \) real roots \( r_1, r_2, \ldots, r_n \), we can write:
   \[
   p(x) = (x - r_1)(x - r_2) \cdots (x - r_n)
   \]
   Since the coefficient of \( x^n \) is 1, the polynomial is monic.

2. **Given conditions:**
   - All roots \( r_i \leq 1 \).
   - \( p(2) = 3^n \).

3. **Express \( p(2) \) in terms of the roots:**
   \[
   p(2) = (2 - r_1)(2 - r_2) \cdots (2 - r_n) = 3^n
   \]

4. **Express \( p(1) \) in terms of the roots:**
   \[
   p(1) = (1 - r_1)(1 - r_2) \cdots (1 - r_n)
   \]

5. **Reformulate the problem using constraints:**
   We need to find the possible values of \( p(1) \) given the constraints:
   \[
   (2 - r_1)(2 - r_2) \cdots (2 - r_n) = 3^n
   \]
   and \( r_i \leq 1 \) for all \( i \).

6. **Use the substitution \( r_i = 2 - x_i \):**
   Let \( x_i = 2 - r_i \). Then \( x_i \geq 1 \) and:
   \[
   x_1 x_2 \cdots x_n = 3^n
   \]
   We need to find the maximum and minimum values of:
   \[
   p(1) = (1 - r_1)(1 - r_2) \cdots (1 - r_n) = (1 - (2 - x_1))(1 - (2 - x_2)) \cdots (1 - (2 - x_n)) = (-1)^n (x_1 - 1)(x_2 - 1) \cdots (x_n - 1)
   \]

7. **Simplify the expression:**
   \[
   p(1) = (-1)^n \prod_{i=1}^n (x_i - 1)
   \]

8. **Apply the AM-GM inequality:**
   By the Arithmetic Mean-Geometric Mean Inequality (AM-GM):
   \[
   \frac{x_1 + x_2 + \cdots + x_n}{n} \geq \sqrt[n]{x_1 x_2 \cdots x_n}
   \]
   Given \( x_1 x_2 \cdots x_n = 3^n \), we have:
   \[
   \frac{x_1 + x_2 + \cdots + x_n}{n} \geq 3 \implies x_1 + x_2 + \cdots + x_n \geq 3n
   \]

9. **Maximize and minimize \( p(1) \):**
   - **Minimum value:** If any \( x_i = 1 \), then \( p(1) = 0 \).
   - **Maximum value:** To maximize \( p(1) \), set \( x_i = 3 \) for all \( i \). Then:
     \[
     p(1) = (-1)^n (3 - 1)^n = (-1)^n \cdot 2^n
     \]

10. **Conclusion:**
    The possible values of \( p(1) \) are \( 0 \) and \( (-1)^n \cdot 2^n \).

The final answer is \( \boxed{ p(1) = 0 } \) or \( p(1) = (-1)^n \cdot 2^n \).