Norton's equivalent circuit with independent sources
I'm assuming all the currents are coming into the node, and I always
  leave Vc at the end of the subtraction (v1-vc, 0-vc, etc.)


For sure you can do that. But notice that the \$I_1\$ current is leaving the \$V_C\$ node.

So the \$I_1\$ current needs to have a different sign. 

I will show you two cases: 

First case: 

I will give a minus sign if the current entering the node (coming into the node).

And the plus sign if the current is leaving the node.

$$-\frac{V_1 - V_C}{R_3} - \frac{0 - V_C}{R_2} - \frac{0 - V_C}{R_4} + I_1 =0$$   

Notice that I give \$I_1\$ the plus sign because \$I_1\$ is leaving the node (current source).   

Second case:

Plus sign to all the currents that are entering the node.

$$\frac{V_1 - V_C}{R_3} + \frac{0 - V_C}{R_2} + \frac{0 - V_C}{R_4} - I_1 =0$$  

Do you know why I give a minus sign to \$ I_1\$? 

So what is important here is that you always must be consistent with the sign convention you have chosen when you are writing nodal equations.

And of course, those two "method" will give the same answer.  

And we do not take \$R_1\$ resistor into account because \$R_1\$ is in series with the constant current source. Hence, it won't change the current in this branch.