Is this normal, or do I have an error somewhere?

Looks normal to me.

Apparently, you copied the first application circuit given on the p.11 of the datasheet. There are some performance graphs on p.13 and onwards. The efficiency graph shows an efficiency of 13~14% for 1 mA output current when the input is 3.7 VDC. This is close to your case for comparison. Now let's work out the input current at this "minimum" load:

$$ \eta=0.13 = \frac{3.3 \cdot 0.001}{3.7 \cdot I_{in}} \Rightarrow I_{in}=6.9 \ \text{mA} $$

1 mA and idle (zero load) are different things but should be enough to give an idea:

For 3.3 mW of output power (3.3V - 1 mA) the circuit consumes roughly 25 mW, according to the performance graphs. Now if we assume the 3.3 mW is transferred to the output with no extra loss then the input power at exactly zero load (idle) would be 25 - 3.3 = 21.5 mW, making the input current roughly 6 mA.

So it looks normal to see 5 mA input current at zero load.

Is this normal, or do I have an error somewhere?

Looks normal to me.

Apparently, you copied the first application circuit given on the p.11 of the datasheet. There are some performance graphs on p.13 and onwards. The efficiency graph shows an efficiency of 13~14% for 1 mA output current when the input is 3.7 VDC. This is close to your case for comparison. Now let's work out the input current at this "minimum" load:

$$ \eta=0.13 = \frac{3.3 \cdot 0.001}{3.7 \cdot I_{in}} \Rightarrow I_{in}=6.9 \ \text{mA} $$

1 mA and idle (zero load) are different things but should be enough to give an idea:

For 3.3 mW of output power (3.3V - 1 mA) the circuit consumes roughly 25 mW, according to the performance graphs. Now let's think straightforward: If we assume the 3.3 mW is transferred to the output with no extra loss then the input power at exactly zero load (idle) would be 25 - 3.3 = 21.5 mW, making the input current roughly 6 mA.

So it looks normal to see 5 mA input current at zero load.