Algebraic way for finding a transfer function of a filter
Your first node equation should be:

$$\frac{\text{V}_\text{1}-\text{V}_{in}}{\text{R}_1}+\frac{\text{V}_1}{\frac{1}{\text{s}\text{C}_1}}+\frac{\text{V}_1-\text{V}_2}{\text{R}_2}=0\tag1$$
Rest are correct:
$$\frac{\text{V}_2-\text{V}_1}{\text{R}_2}+\frac{\text{V}_2-\text{V}_\text{out}}{\frac{1}{\text{s}\text{C}_3}}+\frac{\text{V}_2-\text{V}_3}{\text{R}_3}=0\tag2$$
$$\frac{\text{V}_3-\text{V}_2}{\text{R}_3}+\frac{\text{V}_3}{\frac{1}{\text{s}\text{C}_2}}=0\tag3$$
$$V_3 = V_{out}\tag4$$
We can simplify (3) using (4)
$$\frac{\text{V}_{out}-\text{V}_2}{\text{R}_3}+\frac{\text{V}_{out}}{\frac{1}{\text{s}\text{C}_2}}=0$$
$$\implies V_2 = V_{out}(1+\frac {R_3}{1/sC_2})\tag5$$
You can use (5) to eliminate \$V_2\$ from (1) and (2) to end up in two equations with 3 independent variables of form:
$$f(V_1,V_{in}, V_{out}) = 0 \tag6$$
$$g(V_1,V_{in}, V_{out}) = 0 \tag7$$
You can then find an expression for \$V_1\$ in terms of \$V_{in}\$ and \$V_{out}\$ from (6) as well as from (7). Equate both of them to get a final equation of form:
$$h(V_{in},V_{out}) = 0 $$
You can then sort out \$V_{out}/V_{in}\$ to derive the transfer function.