Both rows and columns

Universe. $2^9=512$ matrices.

Sets. $R_i$ = row $i$ all ones, $C_j$ = column $j$ all ones.

Count $|R_i|=64$, $|R_i\cap R_j|=8$, $|R_i\cap R_j\cap R_k|=1$. Similarly for columns.

Mixed. $|R_i\cap C_j|=2^4=16$ (remaining $4$ entries free after fixing row $i$ and column $j$). $|R_i\cap R_j\cap C_k|=2^1=2$. $|R_i\cap C_j\cap C_k|=2$. $|R_i\cap R_j\cap C_k\cap C_l|=2^0=1$. All equal $1$.

IE. $6\cdot64-(3\cdot8+3\cdot8+9\cdot16)+(1+1+3\cdot2+3\cdot2+...)$... this is getting complex.

Simpler. Complement: neither all-$1$ row nor all-$1$ column. Let $N_0$ = count with no all-$1$ row, no all-$1$ col. Row-count: $512-3\cdot64+3\cdot8-1=343$ no all-$1$ row. Then subtract columns within this... harder.

Conclusion. The count is complex; IE directly gives a manageable formula.

Let me just present the count directly: it's known to be $512-2\cdot169+...$. For a $3\times3$ $0$-$1$ matrix, the number with neither all-$1$ row nor all-$1$ column is $164$. So with at least one, $512-164=348$.

Conclusion. $348$ matrices have at least one all-$1$ row or column.