The suggestion in my comment was wrong -- I was incorrectly (and absurdly) extrapolating from Shioda's examples of inseparably uniruled surfaces (which are beautiful, and which I would love to generalize). Here is the simplest example I know that works in characteristic $0$.
Let $k$ be an algebraically closed field of characteristic $\neq 2,3$, and work in the category of $k$-schemes. Let $\mathbb{P}_k^4$ have homogeneous coordinates $[Z_0,Z_1,Z_2,Z_3,Z_4]$. Let $F$ be the degree $4$ (essentially) Fermat polynomial,
$$
F(Z_0,Z_1,Z_2,Z_3,Z_4) = Z_0^4 + Z_1^4 + Z_2^4 + Z_3^4 + Z_4^4.
$$
Denote by $X\subset \mathbb{P}^4_k$ the zero scheme of $F$. Since $\text{char}(k)\neq 2$, this is a smooth quartic hypersurface of dimension $n=3$. Now let $U$ be the open subset of $X\times X$ that is the complement of the diagonal $\Delta_X$. Let $F\subset U$ be the closed subscheme parameterizing ordered pairs of distinct points $([S_0,S_1,S_2,S_3,S_4],[T_0,T_1,T_2,T_3,T_4])$ such that the homogeneous polynomial in coordinates $[S,T]$,
$$
F(TS_0-ST_0,TS_1,-ST_1,TS_3-ST_3,TS_4-ST_4) =
T^4(S_0^4+S_1^4+S_2^4+S_3^4) - $$
$$
4ST^3(S_0^3T_0+S_1^3T_1+S_2^3T_2+S_3^3T_3+S_4^3T_4) +
$$
$$
6S^2T^2(S_0^2T_0^2 +S_1^2T_1^2 + S_2^2T_2^2 + S_3^2T_3^2 + S_4^2T_4^2) -
$$
$$
4S^3T(S_0T_0^3+S_1T_1^3+S_2T_2^3+S_3T_3^3+S_4T_4^3) +
$$
$$
S^4(T_0^4+T_1^4+T_2^4+T_3^4+T_4^4).
$$
Since the characteristic is not $2$ or $3$, it is straightforward to see that $F$ has codimension $3$ in $U$. It follows that the space of lines, $\overline{\mathcal{M}}_{0,0}(X,1)$ in the stable map notation, is a projective curve.
Now let $i$ be a root of $t^4+1$ in $k$, and consider the hyperplane section $\text{Zero}(Z_4-iZ_3)\cap X$. Using homogeneous coordinates $[Z_0,Z_1,Z_2,Z_3]$ on $\text{Zero}(Z_4-iZ_3)$, the intersection with $X$ is the zero scheme of $Z_0^4+Z_1^4+Z_2^4$. In particular, there is no dependence on $Z_3$. This means that the hyperplane section is a cone with vertex $[Z_0,Z_1,Z_2,Z_3] = [0,0,0,1]$.
Thus the plane curve $$
B= \{[Z_0,Z_1,Z_2] \in \mathbb{P}^2_k |Z_0^4+Z_1^4+Z_2^4 = 0\},
$$
is an irreducible component of the space of lines: for every $[S_0,S_1,S_2]$ in $B$, associate the line $L=\text{span}([S_0,S_1,S_2,0,0],[0,0,0,1,i]\}$. But it is straightforward to compute that the normal bundle of $L$ in $X$ is isomorphic to $\mathcal{O}_L(-2)\oplus \mathcal{O}_L(1)$. Thus, the Zariski tangent space at every point of $B$ is $2$-dimensional. Therefore the space of lines is everywhere reduced along $B$.
I believe this example is an exercise in Debarre's textbook.