Yes, the mass of the observable Universe is always increasing.
Matter
Even if you're only referring the "ordinary" matter (such as stars, gas, and bicycles) and dark matter, the mass of the observable Universe does increase, not because mass is being created, but because the size of the observable Universe increases. In a billion years from now, we can see stuff that today is too far away for the light to have reached us, so its radius has increased. Since the mass $M$ equals density $\rho_\mathrm{M}$ times volume $V$, $M$ increases.
As called2voyage mentions, we have several ways of measuring the density, and we know it's close to $\rho_\mathrm{M}\simeq 2.7\times10^{-30}\,\mathrm{g}\,\mathrm{cm}^{-3}$ (Planck Collaboration et al. 2020). The radius is $R = 4.4\times10^{28}\,\mathrm{cm}$, so the mass is
$$
M = \rho_\mathrm{M} \times V = \rho_\mathrm{M} \times \frac{4\pi}{3}R^3 \simeq 10^{57}\,\mathrm{g},
$$
or $5\times10^{23}M_\odot$ (Solar masses).
Mass increase of matter
Every second, the radius of the observable Universe increases by $dR = c\,dt = 300\,000\,\mathrm{km}$, in addition to the expansion. Here, $c$ is the speed of light, and $dt$ is the time interval that I choose to be 1 second. That means that its mass (currently) increases by
$$
\begin{array}{rcl}
dM & = & A \times dR \times \rho_\mathrm{M}\\
& = & 4\pi R^2 \times c\,dt \times \rho_\mathrm{M}\\
& \sim & 10^6\,M_\odot\,\text{per second,}
\end{array}
$$
where $A=4\pi R^2$ is the surface area of the Universe.
Dark energy
However, another factor contributes to the mass increase, namely the so-called dark energy, which is a form of energy attributed to empty space. And since new space is created as the Universe expands, dark energy is being created all the time. Currently, the energy density of dark energy, expressed as mass density through $E=mc^2$, is more than twice that of matter ($\rho_\Lambda \simeq 6\times10^{-30}\,\mathrm{g}\,\mathrm{cm}^{-3}$).
The rate at which the observable Universe grows due to expansion can be calculated from the Hubble law, which says that objects at a distance $d$ from us recedes at a velocity
$$
v = H_0 \, d,
$$
where $H_0\simeq 70\,\mathrm{km}\,\mathrm{s}^{-1}\,\mathrm{Mpc}^{-1}$ is the Hubble constant. Expansion thus makes the edge of the observable Universe recede at $v=H_0 R = 3.2c$ (yes, more than three times the speed of light), in addition to the factor of $1c$ that comes from more light reaching us (as above).
Mass increase of dark energy
Hence, every second the "total" radius of the observable Universe (i.e. expansion + more light) increases by $dR = (3.2c + 1c)\times dt$, such that the increase in mass/energy from dark energy is
$$
\begin{array}{rcl}
dM & = & A \times dR \times \rho_\Lambda\\
& = & 4\pi R^2 \times (3.2c + 1c)dt \times \rho_\Lambda\\
& \sim & 10^7\,M_\odot\,\text{per second,}
\end{array}
$$
an order of magnitude more than that of regular/dark matter.