Here's an example $G$ for all $i \geq 8$ and $n \geq 2i+1$ that contains only $8$ induced subgraphs (up to isomorphism).

If we choose $i$ vertices from the top row, we obtain $\overline{K_i}$.

If we choose $i-1$ vertices from the top row, we obtain $K_{1,i-1}$ or $\overline{K_i}$.

If we choose $i-2$ vertices from the top row, we obtain $\overline{K_i}$ or $\overline{K_{i-2}} \cup K_2$ or $K_{1,i-2} \cup K_1$ or $K_{1,i-1}$.

If we choose $i-3$ vertices from the top row, we obtain the subgraphs induced by $i-3 \text{ vertices from a,b,c,d,... } \cup \{e,g,h\}$ or $i-3 \text{ vertices from a,b,c,d,... } \cup \{e,f,g\}$ or $K_{1,i-2} \cup K_1$ or $\overline{K_{i-3}} \cup P_3$ (path with three vertices).

If we choose $i-4$ vertices from the top row, we obtain the graph induced by $i-4 \text{ vertices from a,b,c,d,... } \cup \{e,f,g,h\}$.

Since we must choose at least $i-4$ vertices from the top row, in total, that's 8 isomorphism classes of graphs. We can see that $G$ and its complement both contain a triangle (and are therefore not bipartite) and are connected (each vertex is not adjacent to either h or a).

Generalising this technique, we can replace {e,f,g,h} by any non-bipartite connected subgraph of diameter at least 2. In this case, there would exist some I, N such that for all i>=I and n>=max(N,2i+1) which would satisfy the required conditions.