You're trying to use the relay as a current sensor. That's of course possible, but it requires a low voltage relay that has a low coil resistance.
For example, suppose that your sensing relay would be Axicom/TE
IM00 model, with 1.5VDC 16Ω coil. The coil expects about 100mA when operating. Since the current you're sensing is 350mA, you'd need to bypass 250mA around the coil. For that, you can use a resistor with 2.5x the conductance of 16Ω, that is 6.4Ω. Even a lower value, like 5Ω, would work, since the relay cuts-in 75% of the operating voltage. This would further reduce the voltage drop across our current sensor.
Thus, your circuit would be:
simulate this circuit – Schematic created using CircuitLab
D1 and D2 should be close to the coil L1 - ideally mounted right across its terminals. D3 can be any signal/switching diode, e.g. 1N4001 would work just fine. D4 can be most any modern LED. C1-R3 protect the contacts of the pushbutton switch from arcing, prolonging its life.
Note how there's no need for a diode parallel to RL1's coil. Also, the electromagnet's L1 should have two anti-paralleled 12V Zener diodes to equalize the turn-on and turn-off time constants. A single diode has the unfortunate effect of making turn-off up to an order of magnitude slower than turn-on. Sometimes you don't care about it, but sometimes you do.
Now, this circuit isn't just a voltage indicator: it is current detector, and e.g. if the electromagnet coil L1 fails open, you'll get immediate feedback: RL1 will stay open in spite of the button SW1 being closed, and D4 will stay off.
If all you need is a simple indicator without current detection - you don't need any extra relays! Just connect D3-R2-D4 in parallel with the electromagnet coil L1.