35 Ah would be the specified capacity for a "normal" discharge. For a 24V lead battery that would be down to about 20V.
At 1mA, I would be more worried about self-discharge. Excellent batteries claim 2% self-discharge per month, which is 0.7Ah, or the equivalent of 700 hours (about 1 month) of 1mA. The typical number is 5% and it also depends on temperature. At 5%, self-discharge is about 1.7Ah.
Under these conditions you would loose a total of 2.4Ah/month, and your battery would last about 14 months.
However, from 24V to 8V your current consumption may not be constant if your device uses a switching power supply.
A better approach may be to think in terms of power consumption rather than Ah. 1mA at 8V is 8mW, so at 24V that would be about 0.33mA at equal power conversion efficiency.
35 Ah at 24V is approximately 840Wh. At 5% self-discharge, you loose about 42Wh/month for self-discharge and 6Wh/month for your load (8mW.24h/day.30days), about 48Wh/month in total. Which would be about 17 months.
If the 1mA is the current at 24V, then you consume 24mW or 17Wh/month. In that case the estimation would be 840Wh/((42Wh+17Wh)/month)=14 months which is very close to the first estimate.
Estimating the duration using the "Ah method" is generally good enough as the voltage variation has to be kept within reasonable limits which has in turn a limited impact on the power consumption itself.
If you are willing to go down to the 8V limit, you should know that the voltage drops very fast at the end of charge and you would not be getting a lot more energy out of the battery anyway. For a very slow discharge, the discharge graphs are almost vertical near the end of life.
One final remark: if your lead battery has a voltage of 24V, then it is no longer fully charged and it is in practice near it's end of charge. There would be about 20% left. So if you measure 24V at the battery terminals you have to divide all numbers above by 5 and your battery would last about 3 months under nominal conditions (20°C).