We have $x^2 + 7 = x^2 + 2^2 + 1^2 + 1^2 + 1^2$. Below we prove that $n = 4$ is impossible (and hence smaller $n$ is also impossible, since we can take $f_i = 0$).

Suppose for contradiction that $x^2 + 7 = f_1(x)^2 + f_2(x)^2 + f_3(x)^2 + f_4(x)^2$ where all are polynomials with rational coefficients. Clearly the $f_i$ must all be of degree one, write $f_i = a_i x + b_i$, $i = 1, 2, 3, 4$. We have
$$\sum_{i=1}^{4} a_{i}^{2} = 1, \quad \sum_{i=1}^{4} a_{i}b_{i} = 0, \quad \sum_{i=1}^{4} b_{i}^{2} = 7.$$
Let $p_i = a_i + b_i$, $q_i = a_i - b_i$, then computation gives
$$\sum_{i=1}^{4} p_{i}^{2} = 8, \quad \sum_{i=1}^{4} q_{i}^{2} = 8, \quad \sum_{i=1}^{4} p_{i}q_{i} = -6.$$
After clearing denominators, this makes the following proposition hold: Proposition: There exist a positive integer $m$ and integers $x_i, y_i (i = 1, 2, 3, 4)$ such that
$$\sum_{i=1}^{4} x_{i}^{2} = 8m^{2}, \quad \sum_{i=1}^{4} y_{i}^{2} = 8m^{2}, \quad \sum_{i=1}^{4} x_{i}y_{i} = -6m^{2}.$$
But we will next prove that the proposition is false, thereby deriving a contradiction. Suppose that the current $m$ is the smallest $m$ for which the proposition holds. Since the square of an odd number leaves remainder $1$ upon division by $8$, and the square of an even number leaves remainder $0$ or $4$ upon division by $8$, a simple case analysis shows that if the sum of four squares is a multiple of eight, then all four numbers must be even, hence $x_i, y_i$ are all even.

But then $-6m^2 = \sum_{i=1}^{4} x_i y_i$ would be a multiple of 4, so $m$ is even, and therefore we can take half of each of $x_i, y_i$ and $m$ and still satisfy the proposition, but this contradicts the assumption that $m$ is the smallest. This completes the proof.