1. **Identify the possible values of $x_i^2 \mod 5$:**
Since $x_i$ is chosen from the set $\{1, 2, 3, 4, 5\}$, we compute $x_i^2 \mod 5$:
$$\begin{aligned} 1^2 &\equiv 1 \pmod{5}, \\ 2^2 &\equiv 4 \equiv -1 \pmod{5}, \\ 3^2 &\equiv 9 \equiv 4 \equiv -1 \pmod{5}, \\ 4^2 &\equiv 16 \equiv 1 \pmod{5}, \\ 5^2 &\equiv 25 \equiv 0 \pmod{5}. \end{aligned}$$
Therefore, $x_i^2 \mod 5 \in \{0, 1, -1\}$.

2. Calculate the probabilities:
$$\mathbb{P}(x_i^2 \equiv 1 \pmod{5}) = \mathbb{P}(x_i^2 \equiv -1 \pmod{5}) = \frac{2}{5}, \quad \mathbb{P}(x_i^2 \equiv 0 \pmod{5}) = \frac{1}{5}.$$

3. **Define the sum $S_n$ and probabilities $p_n, q_n, r_n$:**
Let $S_n = \sum_{i=1}^n x_i^2$. Define:
$$\begin{aligned} p_n &= \mathbb{P}(S_n \equiv 0 \pmod{5}), \\ q_n &= \mathbb{P}(S_n \equiv 1 \pmod{5}), \\ r_n &= \mathbb{P}(S_n \equiv 2 \pmod{5}). \end{aligned}$$

4. Key observation:
If $S_n \equiv 1 \pmod{5}$, then $\sum_{i=1}^n (2x_i)^2 \equiv -1 \pmod{5}$. This implies:
$$\mathbb{P}(S_n \equiv -1 \pmod{5}) = q_n.$$
Similarly, $\mathbb{P}(S_n \equiv -2 \pmod{5}) = r_n$.

5. Set up the recursion relations:
Using the law of total probability with respect to $x_n$:
$$\begin{aligned} p_n &= \frac{1}{5} p_{n-1} + \frac{4}{5} q_{n-1}, \\ q_n &= \frac{2}{5} p_{n-1} + \frac{1}{5} q_{n-1} + \frac{2}{5} r_{n-1}, \\ r_n &= \frac{2}{5} q_{n-1} + \frac{3}{5} r_{n-1}. \end{aligned}$$

6. Matrix form of the recursion:
$$\begin{pmatrix} p_n \\ q_n \\ r_n \end{pmatrix} = \begin{pmatrix} \frac{1}{5} & \frac{4}{5} & 0 \\ \frac{2}{5} & \frac{1}{5} & \frac{2}{5} \\ 0 & \frac{2}{5} & \frac{3}{5} \end{pmatrix} \begin{pmatrix} p_{n-1} \\ q_{n-1} \\ r_{n-1} \end{pmatrix}.$$

7. Initial conditions:
$$(p_1, q_1, r_1) = \left(\frac{1}{5}, \frac{2}{5}, 0\right).$$

8. Analyze the recursion:
By studying the recursion, we find:
$$p_n = \frac{1}{5} \text{ for every odd } n, \quad p_n > \frac{1}{5} \text{ for every even } n.$$
This can be shown by induction, considering the parity of $n$.

Conclusion:
$$\boxed{\text{The probability that } S_n \equiv 0 \pmod{5} \text{ is at least } \frac{1}{5}.}$$