= Solution
Write $h_{ij}={}^{(3)}g_{ij}$ for the <induced metric> and lower the <shift vector> with it, $N_i=h_{ij}N^j$. In the stated negative-shift convention the four-dimensional components and inverse components are
$$
g_{00}=-N^2+N_iN^i,\quad g_{0i}=-N_i,\quad g_{ij}=h_{ij},
\qquad g^{00}=-\frac1{N^2},\quad g^{0i}=-\frac{N^i}{N^2}.
$$
Multiplying the given future <unit normal> by this <metric tensor> gives $n_\mu=(-N,0,0,0)$. In particular the tangential normal components vanish, but their <covariant derivatives> do not:
$$
n_{i;j}=-\Gamma^\mu{}_{ij}n_\mu=N\Gamma^0{}_{ij},\qquad K_{ij}=-N\Gamma^0{}_{ij}.
$$
The relevant <Christoffel symbol> is
$$
\begin{aligned}
\Gamma^0{}_{ij}
&=\frac1{2N^2}\left[\dot h_{ij}+\partial_iN_j+\partial_jN_i
-N^k(\partial_i h_{jk}+\partial_jh_{ik}-\partial_kh_{ij})\right]\\
&=\frac1{2N^2}\left(\dot h_{ij}+N_{i|j}+N_{j|i}\right).
\end{aligned}
$$
For the second equality, the last parenthesis is $2h_{k\ell}{}^{(3)}\Gamma^\ell{}_{ij}$, so its contraction subtracts the two spatial connection terms in the symmetrized derivative of $N_i$. Therefore
$$
\boxed{K_{ij}=-\frac1{2N}\left(\dot h_{ij}+N_{i|j}+N_{j|i}\right).}
$$
This is <extrinsic curvature with a negative shift>. Both the future-normal orientation and the definition $K=-\nabla n$ matter: replacing the threading by $dx^i+N^idt$ reverses the shift terms. The spatial <induced metric> is generally a function of time as well as position; restricting it to time-independent data would simply make the displayed time derivative vanish.
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