Distance between SPI traces to prevent cross talk
The main source of crosstalk will presumably be capacitive coupling between the traces (though see analogsystemrf's answer for an analysis of inductive coupling). Suppose we have a pair of traces, 0.1 mm apart (which is about the closest you'll find on a standard PCB) and 35 μm thick (i.e. 1 oz copper). This calculator claims that the resulting capacitance will be about 1.2 pF for a 10 cm trace.

Suppose you run the SPI bus at 24 MHz. At this frequency, the capacitor will have an impedance of \$\frac{1}{2\pi\cdot f\cdot C} \approx 5.5~\text{k}\Omega\$. However, a square wave has harmonics, multiples of the base frequency. An ideal square wave has only odd harmonics, and can be represented as $$\sum_{k=1}^\infty \frac{1}{2k-1}\sin\big((2k-1)\cdot\omega t\big)$$ In practice, you can get a decent square wave with only through the fifth harmonic, $$\sin(\omega t) + \frac{1}{3}\sin(3\omega t) + \frac{1}{5}\sin(5\omega t)$$ At the higher frequencies, the impedance of the capacitor will be proportionately lower, but the voltage of those harmonics is also proportionately lower, so each sinusoid contributes about equally to the crosstalk.

Compare the impedance of the capacitor at the fundamental frequency to the approximately 50 Ω output impedance of a microcontroller GPIO pin, and we can see that there is attenuation of about a factor of 110. With contributions from the fundamental, third, and fifth harmonics, the crosstalk is about a factor of about 36 below the signal. 

To be able to make the approximation that we only care about the fifth harmonic and below, we need to make sure that the signal is bandwidth-limited, which requires low-pass filtering. The inherent capacitance of the traces and input pin is maybe 12 pF, which with the 50 Ω driver impedance forms a low pass filter with a -3db frequency of \$\frac{1}{2\pi\cdot R\cdot C} \approx 265~\text{MHz}\$. This is a bit high to cut off the seventh harmonic of 168 MHz, so you can add a bit more resistance or capacitance to the trace if crosstalk becomes a problem, though I'd be surprised if it does at these frequencies and distances.