The problem with your circuit is that the base currents for all of the transistors flows through the D1 string, as well as the collector current for Q1.
User2233709's answer shows a good, and commonly used solution to that problem. That is, the use of an "emitter follower augmented current mirror" or a "buffered feedback current mirror". The idea is that the base current is separated from the reference current.
Unfortunately, the PCB for this circuit has already been manufactured and assembled. Hence, the only modifications I can make are by replacing components and not adding anything.
This poses a serious problem. :-(
What needs to be done is to change both the emitter resistor of Q1 and Q1 itself.
To decrease the current through the D1 line without decreasing the current through the other lines, you need to
- decrease the current through the collector of Q1
- maintain the same voltage at the bases of the transistors, and hence the collector of Q1, to which the bases are tied.
The difficulty is in choosing a transistor that will be a suitable substitute.
Since you want 60 mA through all the lines, and the emitter resistors for the lines other than D1 are 8.7 \$\Omega\$, there is about 522 mV across each emitter resistor.
Since you say the gain of the transistors is about 50, and there is 60 mA through their collectors, the base current for each transistor is about 1.2 mA. You have 40 strings, so the total base current is about 48 mA. So, ideally, you want the current through the reference transistor's collector to be only about 12 mA.
(I think you can already see that errors in estimations will pose problems).
12 mA is 1/5 of 60 mA, so to maintain the 522 mV across the emitter resistor, the R2 must be increased to 8.7 x 5 = 43.5 \$\Omega\$.
The \$V_{BE} : I_C\$ relationship for a diode connected transistor is given by
$$I_C = I_S( e^{\frac{V_{BE}}{V_T}} -1 ) \approx I_S e^{\frac{V_{BE}}{V_T}}$$
where \$I_S\$ is the reverse saturation current.
To reduce \$I_C\$ to 1/5th it's value, you need to find a transistor which has an \$I_S\$ 1/5th that of the BF720.
Unfortunately, but not surprisingly, the datasheet of the BF720 does not list a value for \$I_S\$. Nor does it have a graph or datapoints for the \$V_{BE} : I_C\$ relationship from which one might compute \$I_S\$.
If one has really good SPICE models of the BF720 and other transistors it might be possible to find a transistor that, in theory at least, has an \$I_S\$ 1/5th that of a BF720. Or one might choose a transistor that has 1/5th of the rated current of a BF720, in the hopes that the current rating is approximately proportional to \$I_S\$. But in the end, one must test such a transistor against a BF720 to determine if it really does conduct 1/5th the current when diode connected, at the same \$V_{BE}\$.
Success is far from guaranteed. :-( Sorry for the bad news.
As @mkeith has suggested, you might try replacing all of the transistors with MOSFETs. Mosfet current mirrors have different problems, but at least you will not have the problem of too much base (gate) current flowing through the reference line.