It's more convenient to start the sum at $i=0$ instead of $i=1$,
adding the term ${0 \choose 0} = 1$ and aiming for
$$
\sum_{i=0}^{(p-1)/2} {2i \choose i} \equiv \pm 1 \bmod p.
$$
To prove this, note that ${2i \choose i} \equiv 0 \bmod p$ for
$(p-1)/2 < i \leq p-1$. Therefore
$$
\left(
\sum_{i=0}^{(p-1)/2} {2i \choose i}
\right)^2 =
\sum_{i=0}^{(p-1)/2} \, \sum_{j=0}^{(p-1)/2} {2i \choose i} {2j \choose j}
\equiv
\mathop{\sum\sum}_{i,j \geq 0}^{i+j \leq p-1} {2i \choose i} {2j \choose j}
\bmod p.
$$
But $\sum_{i+j = n} {2i \choose i} {2j \choose j} = 4^n$
because $2i\choose i$ is the $x^i$ coefficient of $(1-4x)^{-1/2}$.
Hence
$$
\left(
\sum_{i=0}^{(p-1)/2} {2i \choose i}
\right)^2 \equiv \sum_{n=0}^{p-1} 4^n
= 1 + \sum_{n=1}^{p-1} 4^n \bmod p.
$$
Since we assumed $p \geq 5$, the last sum is
$4 \cdot (4^{p-1}-1)/(4-1) \equiv 0 \bmod p$ (by Fermat little theorem), so
$\sum_{i=0}^{(p-1)/2} {2i \choose i}$ is a square root of $1 \bmod p$,
QED.
barak manos observes in a comment that the sign $1$ or $-1$ seems to match
the residue of $p \bmod 6$ (equivalently, of $p \bmod 3$ because $p$ is odd).
I have checked this experimentally for all $p < 1000$. Proving it
will likely require a different approach.
Postscript not so different, as it turns out.
Consider the polynomial
$$ P(x) = \sum_{i=0}^{(p-1)/2} {2i \choose i} (x/4)^i \bmod p. $$
The same argument as before shows that
$P(x)^2 \equiv (x^p - 1) / (x-1) \bmod p$ identically;
but that means $P(x)^2 \equiv (x-1)^{p-1}$, so
$P(x) = \pm (x-1)^{(p-1)/2}$ (which can then also be seen directly).
Comparing the coefficients of $x^{(p-1)/2}$ (or more simply, of $x$)
we soon see that the sign is the Legendre symbol $(-1/p)$.
Substituting some $x \bmod p$ into $(x-1)^{(p-1)/2}$
yields the Legendre symbol $((x-1)/p)$.
Here $x=4$ so we get $(-1/p) (3/p) = (-3/p)$, which by quadratic reciprocity
is $1$ or $-1$ according as $p$ is $1$ or $-1 \bmod 3$, QED.
