First, take the case without encoding. Let there be an $X$ error with probability $p$, so the output state is
$$
\rho=(1-p)|\psi\rangle\langle\psi|+pX|\psi\rangle\langle\psi|X.
$$
Normally, I would just talk about the probability of an error (it lets you avoid some of the averaging I'm about to do), but we can talk about fidelity if you want.
$$
F=\langle\psi|\rho|\psi\rangle=1-p+p|\langle\psi|X|\psi\rangle|^2
$$
To get just a single number, you might average over all possible initial states $|\psi\rangle$. You'll find that
$$
\bar F=1-\gamma p
$$
for some positive value $\gamma$ that is not so important to calculate right now.
By comparison, consider the encoded qubit going through the error channel. If
$$
|\Psi\rangle=\alpha_0|000\rangle+\alpha_1|111\rangle,
$$
then
$$
\rho_2=(1-p)^3|\Psi\rangle\langle\Psi|+(1-p)^2p(X_1|\Psi\rangle\langle\Psi|X_1+X_2|\Psi\rangle\langle\Psi|X_2+X_3|\Psi\rangle\langle\Psi|X_3)+(1-p)p^2X_1X_2|\Psi\rangle\langle\Psi|X_1X_2+\ldots+p^3X_1X_2X_3|\Psi\rangle\langle\Psi|X_1X_2X_3.
$$
If you error correct this, it will succeed so long as there's no more than one error
$$
\rho_2\rightarrow (1-p)^2(1+2p)|\Psi\rangle\langle\Psi|+p^2(3-2p)X_1X_2X_3|\Psi\rangle\langle\Psi|X_1X_2X_3
$$
which will decode to
$$
(1-p)^2(1+2p)|\psi\rangle\langle\psi|+p^2(3-2p)X|\psi\rangle\langle\psi|X
$$
and hence have fidelity
$$
\bar F=1-\gamma p^2(3-2p).
$$
This fidelity is better than the fidelity from no encoding provided
$$
p^2(3-2p)<p.
$$
This is fulfilled for $0<p<\frac12$.
